TEACH YOURSELF VISUALLY - CALCULUS

9Additional Applications of Differentiation: Word Problems chapter

2 Relate w and L using the red and blue similar right triangles. 12 = 5
w+ L
L

3 Solve for L in terms of w. 12L = 5^w + Lh

12L = 5w + 5L

7L = 5w

L = 5 w
7

4 Differentiate with respect to time, t. dL = 5 dw
dt 7 dt

5 Substitute dw = 4 and then solve for dL . dL = 5 $ 4
dt dt dt 7

dL = 20
dt 7

The length of her

shadow is changing at

a rate of 20 ft./ sec.
7

187

Position, Velocity,
and Acceleration

In previous chapters, you found that if s(t) = the position function, then s'(t) = v(t) = the velocity
function and s"(t) = v'(t) = a(t) = the acceleration function.

Rocket Problem

A small toy rocket is shot into the air from the top of a tower. Its position, s, in feet, t seconds after
liftoff, is given by the equation s(t) = –16t2 + 32t + 240.

PART A s(t) = –16t2 + 32t + 240
Find the velocity of the rocket 2 seconds after liftoff. s'(t) = v(t) = –32t + 32

1 Starting with the position function s(t), find the velocity
function v(t).

2 Find v(2). v(t) = –32t + 32
v(2) = –32(2) + 32
v(2) = –32

After 2 seconds, the rocket’s velocity is –32
feet/second. (The negative velocity indicates
that the rocket is moving downward.)

PART B s ^th = - 16t 2+ 32t+ 240
For how many seconds will the rocket be in the air? 0 = - 16t 2+ 32t+ 240
Find the time at which the rocket lands on the ground—that is, 0 = - 16 _t 2+ 2t- 15i
when s(t) = 0. 0 = - 16^t - 5h^t+ 3h
t = 5 t= - 3
188
The rocket is in the air
for 5 seconds.

9Additional Applications of Differentiation: Word Problems chapter

PART C v(t) = –32t + 32
With what velocity will the rocket hit the ground?

1 Start with the velocity function v(t).

2 Find v(5). v(5) = –32 • 5 + 32
v(5) = –160 + 32
Note: We used t = 5, since that’s the time v(5) = –128
when the rocket hits the ground.
The rocket hits the ground with a velocity of
–128 feet/second. Again, the negative indicates
that the rocket is moving downward.

PART D v(t) = –32t + 32
How many seconds after liftoff will the rocket reach its 0 = –32t + 32
maximum height?
–32t = 32
You need to find when the rocket stops moving—that is, t=1
when v(t) = 0.
The rocket reaches its maximum
height after 1 second.

PART E s(t) = –16t2 + 32t + 240
Find the maximum height reached by the rocket.

1 Start with position function, s(t).

189

Position, Velocity, and s(1) = –16(1)2 + 3291) + 240
Acceleration (continued) s(1) = –16 + 32 + 240
s(1) = 256
2 Find s(1).
The maximum height reached
by the rocket is 256 feet.

Particle Moving Along a Straight Line Problem

This type of problem is also known as a “rectilinear motion” problem. A particle moves along the x-axis
so that its x-coordinate at time t (seconds) is given by the position function: x(t) = 3t4 – 28t3 + 60t2.

PART A x(t) = 3t4 – 28t + 60t2
At what time is the particle at rest? x'(t) = v(t) = 12t3 – 84t2 + 120t

1 The particle is at rest when its velocity is 0. First, find
the velocity function.

2 Set v(t) = 0 and then solve for t. 0 = 12t3 – 84t2 + 120t
0 = 12t(t2 – 7t + 10)
0 = 12t(t – 2)(t – 5)
t = 0, t = 2, and t = 5

The particle is at rest at 0, 2, and
5 seconds.

190

9Additional Applications of Differentiation: Word Problems chapter

PART B x' (t ) = v(t) = 12t3 – 84t2 + 120t
During what time intervals is the particle
moving to the left? During what time intervals 0<t<2 2<t<5 t>5
is the particle moving to the right?
v(1) > 0 v(3) < 0 v(6) > 0
Create a first derivative chart, using the
numbers obtained in Part A. → ←→

TIP right left right

When the velocity is negative, the particle is The particle is moving right when 0 < t < 2
moving left. and when t > 5.

When the velocity is positive, the particle is The particle is moving left when 2 < t < 5.
moving right.

PART C x(t) = 3t4 – 28t3 + 60t2
Find the total distance traveled by the particle in the first 5 seconds.

1 Start with the position function, x(t).

2 Using the zeros of v(t), t = 0, t = 2, and t = 5, find the value of x(0) = 0
x(t) for each of these times. x(2) = 64
x(5) = –125

3 Find the distance traveled in each time From 0 to 2 seconds: 64 – 0 = 64 units
interval, and then find the sum of these
two distances. From 2 to 5 seconds: 64 – (–125) = 189 units

Total distance traveled in first 5 seconds is
64 + 189 = 253 units.

191

Position, Velocity, and v(t) = 12t3 – 84t2 + 120t
Acceleration (continued)

PART D
What is the particle’s acceleration at t = 1?

1 Start with the velocity function v(t).

2 Find a(t), the acceleration function. v'(t) = a(t) = 36t2 – 168t + 120

3 Find a(1). a(1) = 36(1)2 – 168(1) + 120
a(1) = –12

The particle’s acceleration at t = 1 is –12 ft/sec2,
with the negative indicating that the particle is
slowing down or “decelerating.”

PART E a(t) = 36t2 – 168t + 120
At what time is the particle moving with constant velocity?

1 Begin with the acceleration function, a(t).

192

9Additional Applications of Differentiation: Word Problems chapter

2 The particle moves at constant velocity when it is not 0 = 36t2 – 168t + 120
accelerating, so set a(t) = 0. 0 = 12(3t2 – 14t + 10)

3 Since the equation factors no further, use the -^-14h ! ^-14h2- 4 ^3h^10h
Quadratic Formula to find the values of t. t = 2 ^3h

t = 14 ! 196 - 120
6

t = 14 ! 2 19
6

t = 7 ! 19
3

The particle is moving with constant
velocity at approximately 3.79 seconds
and 0.88 seconds.

TIP

For the quadratic equation ax2 + bx + c = 0, the
solution is

- b ! b 2– 4ac
x= 2a

193

10chapter

Introduction to
the Integral

This chapter is the first of three chapters that deal Antiderivatives: Differentiation
with the process of starting with the derivative versus Integration . . . . . . . . . . . . . . . 195
of a function and working backward to get the
original function, called the antiderivative. This The Indefinite Integral and
process is known as integration. This chapter Its Properties . . . . . . . . . . . . . . . . . . . 197
covers both the indefinite and definite integrals,
along with their properties, as well as the First and Common Integral Forms . . . . . . . . . . . 201
Second Fundamental Theorems of Calculus and the
Mean Value Theorem. First Fundamental Theorem
of Calculus . . . . . . . . . . . . . . . . . . . . . 203

The Definite Integral and Area . . . . . . 205

Second Fundamental Theorem of
Calculus. . . . . . . . . . . . . . . . . . . . . . . . 209

Antiderivatives: Differentiation 10chapter
versus Integration

The process of finding a function from which a given derivative is derived is known as
antidifferentiation, or integration. This section introduces that relationship and covers the
indefinite integral and its properties.

Definition of an Antiderivative

A function, F, is called an antiderivative of function f on an interval if F'(x) = f(x) for all x in that
interval.
Let F(x) = x3 – 7x + 6; then F '(x) = f(x) = 3x2 – 7.

1 One antiderivative of f(x) = 3x2 – 7 is the function F(x) at right. F(x) = x3 – 7x + 6

2 A second antiderivative of f(x) = 3x2 – 7 is the function F(x) = x3 – 7x – 15
F(x) at right.
You, therefore, write the most general
3 In each case above, F '(x) = f(x). So it appears antiderivative of f(x) = 3x2 – 7 as
that a given function f(x) has an infinite number F(x) = x3 – 7x + c, where c is just
of antiderivatives, F(x), all differing from each some constant.
other by just a constant.

195

Antiderivatives: Differentiation
versus Integration (continued)

Finding Some Antiderivatives F(x) = sin x + c because F'(x) = f(x)
1 Find the antiderivative of f(x) = cos x.

2 Find the antiderivative of f ^xh = 1x. F(x) = ln x + c because F'(x) = f(x).

3 Find the antiderivative of f(x) = ex. F(x) = ex + c because F '(x) = f(x).

196

The Indefinite Integral 10Introduction to
and Its Properties the Integral chapter

This section introduces the indefinite integral (an antiderivative) of a function along with its
properties. The section also includes some examples of finding indefinite integrals.

Definition of the Indefinite Integral

The indefinite integral of a function f(x), written as ͐ f ^xh dx, is the set of all antiderivatives of the

function f(x).

In the expression ͐ f ^xh dx: ͐ f ^xh dx is read
͐ is the integral symbol.
“the integral of f of
“f(x)” is called the integrand. x with respect to x.”
“dx” tells you that the variable of integration is x.

FINDING SOME INDEFINITE INTEGRALS

1 Because d _ x 3 - 7xi = 3x 2 - 7 & ͐ _3x2- 7i dx = x3- 7x + c
dx

2 Since d ^sin xh = cos x & ͐ cos xdx = sin x + c
dx ͐ _2e2xi dx = e2x+ c

3 Because d _e 2x i = 2e2x &
dx

197

The Indefinite Integral and
Its Properties (continued)

4 Since d ^ln xh = 1 & ͐ b 1 l dx = ln x + c
dx x x

TIP

The x is there since you can’t take the
natural log of a negative number.

FINDING A PARTICULAR ANTIDERIVATIVE
Find the particular antiderivative of f '(x) = 3x2 – 7 that satisfies the condition f(1) = 3.

You need to find a specific or “particular” value of c for the antiderivative of 3x2 – 7.

1 f ^xh = ͐ f l^xh dx f ^xh = ͐ _3x2- 7i dx

f ^xh = x3- 7x + c

This was shown in the
preceding section.

2 Since f(1) = 3, when x = 1, f(x) = 3. 3 = (1)3 – 7(1) + c
9=c

3 Since you found a particular (or specific) value of c, f(x) = x3 – 7x + 9 is the particular
antiderivative of f '(x) = 3x2 – 7.
you have found a “particular” antiderivative of
f '(x) = 3x2 – 7.

TIP

Continuity Implies Integrability
If a function f is continuous on the closed interval [a,b], then f is also integrable
on [a,b]. (The term integrable means that you are able to integrate it.)

198

10Introduction to the Integral chapter

FINDING A FUNCTION FROM ITS SECOND DERIVATIVE
Find the function f(x) for which f"(x) = 32, f '(1) = 36, and f(1) = 16.

1 Find f '(x) from the given data f "(x) = 32. f l^xh = ͐ f m^xh dx
f l^xh = ͐32dx

f l^xh = 32x + c1

2 You are given f '(1) = 36; use this to find the value of c1. f '(1) = 32(1) + c1
36 = 32 + c1
4 = c1

Therefore, f '(x) = 32x + 4.

3 Find f(x). f ^xh = ͐ f l^xh dx
f ^xh = ͐ ^32x+ 4h dx
Note: A second constant, c2, is used here. The constant c1 is
from the first integration step. We cannot assume that these are f ^xh = 16x2+ 4x + c2
equal, so they need to be labeled separately.

4 You are told that f(1) = 16; use this fact to find the f '(1) = 16(1)2 + 4(1) + c2
value of c2. 16 = 16 + 4 + c2
–4 = c2

Thus, f(x) = 16x2 + 4x – 4.

199

The Indefinite Integral and
Its Properties (continued)

Properties of Indefinite Integrals

If f and g are continuous functions and defined on the same interval and K is some constant, then the
following properties apply:

1 Integral of dx: ͐dx = x+ c ͐5dx = 5x + c
2 Integral of a constant: ͐k dx = k x + c
͐ 1 dx = 1 x + c
2 2

3 Integral of constant times a function: ͐4 _3x2- 7i dx= 4͐ _3x2- 7i dx

͐k $ f ^xh dx = 5 $ ͐ f ^xh dx , where 4c = m is just = 4 _x3- 7x + ci
= 4x3- 28x + 4c
another constant. = 4x3- 28x + m

Another way to deal with the constant is as follows: ͐4 _3x2- 7i dx= 4͐ _3x2- 7i dx

= 4 _x3- 7xi + c
= 4x3- 28x + c
Just integrate everything and put a
+ c at the end.

4 Integral of the sum/difference of functions: ͐ bcos x + 1 l dx= ͐ cos xdx + ͐ 1 dx
x x
͐ _ f ^xh ! g^xhi dx = ͐ f ^xh dx ! ͐g^xh dx
= sin x+ ln x + c

Note that only one “+ c” was written; if
you used separate “+ constant” for each
function, their sum would just be another
constant anyway.

200

Common 10Introduction to the Integral chapter
Integral Forms

To create an integral formula from a known derivative formula, just write the formula in

“reverse,” adding the correct integral notation and the “+ c.” For example, since d ^sin xh = cos x,
dx
you can also write that ͐ cos xdx = sin x + c. The following integral formulas were created by just

reading an existing differentiation formula in reverse.

POWER ͐x3 dx = x4 + c or 1 x4+ c
4 4
Use the formula below to R V
integrate some power of a S W
variable. If you were to ͐ 1= ͐ 1 x- 1/2 dx= 1 S x1/2 W+ c= x1/2+ c= x+ c
differentiate the right side, x dx 2 2 SS WW
you would end up with the 2 T 1 X
left side. 2

͐xn dx = xn+ 1 + c, for n ! -1
n+
1

POLYNOMIAL ͐ _3x2- 6x+ 5i dx

Using a combination of the properties listed in previous = 3 $ x3 - 6 $ x2 + 5x + c
sections and the Power Rule listed above, you can find the 3 2
integral of a polynomial as follows:
= x3- 3x2+ 5x + c

͐ _an x n+ an- 1+ f + a2 x2+ a1 x1+ a0i dx

= an xn+ 1 + an- 1 xn + fa2 x3 + a1 x2 + a0 x + c
n+ n 3 2
1

201

Common Integral
Forms (continued)

NATURAL LOGARITHM ͐ 3 dx = 3͐ 1 dx= 3 ln x + c
x x
1
Integrating the expression x is just a matter of using the Note: Since you cannot find
the natural log of a negative
derivative on lnx in reverse. Since d ln x = 1 the number, the absolute value
dx x ⏐n⏐x⏐ is used.

following formula must be true.

͐ 1 dx = ln x + c
x

EXPONENTIAL ͐5ex dx = 5͐ex dx= 5ex+ c

The following derivative rules come from their appropriate ͐2x dx = 2x + c
derivative counterparts for exponential functions found in ln 2
Chapter 6.

͐ex dx = ex+ c and ͐ax dx = ax + c
ln a

TRIGONOMETRIC: COSINE AND SINE ͐ ^2 cos x + 3 sin xh dx
The following integral formulas follow directly from their = ͐2 cos xdx + ͐3 sin xdx
derivative counterparts found in Chapter 5. = 2͐ cos xdx + 3͐ sin xdx

͐ cos xdx = sin x + c = 2 sin x + 3^- cos xh + c
͐ sin xdx = - cos x + c = 2 sin x - 3 cos x + c

TRIGONOMETRIC: SOME OF THE OTHERS ͐ _sec2 x + sec x tan xi dx
Looking back at the trigonometric derivative formulas in = ͐ sec2 xdx + ͐ sec x tan xdx
Chapter 6, you can see that the following integral formulas
result from the process of antidifferentiation. = tan x + sec x + c
= tan x + sec x + c
͐ sec2 xdx = tan x + c
͐ csc2 xdx = - cot x + c
͐ sec x tan xdx = sec x + c
͐ csc x cot xdx = - csc x + c

202

First Fundamental 10Introduction to the Integral chapter
Theorem of Calculus

If the function f is continuous on the closed interval [a,b] and F is an antiderivative of f (that is,

F'(x) = f(x) ) on the interval [a,b], then ͐b f ^xh dx = F ^bh- F ^ah.
a

b

Another way to write the final result is = 8F ^xhB = F ^bh - F ^ah . In other words, it says “after finding
a

the antiderivate, F(x), find the value at the top limit of integration, F(b), then find the value at the bottom

limit of integration, F(a), and then find their difference, F(b) – F(a).”

The a and b on the integral sign ͐b f ^xh dx are called the limits of integration, and the dx indicates that
a

a and b are x values; thus the function f(x) being integrated must be a function of x.

The expression ͐b f ^xh dx is called a definite integral.
a

Note: The definite integral ͐b f ^xh dx represents a number (a definite value), while the indefinite
a

integral f(x)dx represents a family of functions (remember the “+ c”), and not a definite, or specific,

function.

EVALUATE A DEFINITE INTEGRAL: ͐b b
EXPONENTIAL FUNCTION a
f ^xh dx = 8F ^xhB = F ^bh - F ^ah
Evaluate ͐1 ex dx. a
0
follow the pattern above for problem below

͐1 ex dx = 7e x 1 = e1- e0= e - 1
0
A 0

EVALUATE A DEFINITE ͐ π/6 cos xdx = 6sin x @π/6 = sin b π l - sin ^0h = 1 - 0 = 1
INTEGRAL: TRIGONOMETRIC 0 0 6 2 2
FUNCTION

Evaluate ͐π/6 cos xdx.
0

203

First Fundamental Theorem
of Calculus (continued)

EVALUATE A DEFINITE INTEGRAL: POLYNOMIAL FUNCTION

Evaluate ͐2 _2x - 3x2i dx.
-1

1 Find the integral with limits written on the bracket. ͐2 _2x - 3x 2i dx
-1

= 7 x2- x3A2
-1

2 With F(x) = x2 – x3, find F(2) – F(–1). = _22- 23i - a^-1h2- ^-1h3k

\ 1 4 44 2 4 44 3
F ^2h F ^-1h

= ^4 - 8h -^1 + 1h

= -6

204

The Definite Integral
10and Area
Introduction to the Integral chapter

One of the applications of the definite integral is finding the y
area of a region bounded by the graphs of two functions. y = f(x)

Let f be a continuous function on [a,b] for which f(x) ≥ 0 for all x

in [a,b]. Let R be the region bounded by the graphs of y = f(x) and

the x-axis and the vertical lines x = a and x = b. Then the area, AR, R

of the region is given by ͐b f ^xh dx . x

AR= a

a b
x=a x=b

Area of a Bounded Region: Linear Function y

Find the area of the region bounded by the graph of y = 2x, y = 0 y = 2x
(the x-axis), and the lines x = 0 (the y-axis) and x = 3.
1 Sketch a diagram of the bounded region.

R

0 3 y=0
x=0 x=3

2 Set up the integral with a = 0, b = 3, and f(x) = 2x. ͐b f ^xh dx

AR= a

= ͐3 2xdx
0

205

The Definite Integral 2x 2 3 2 3 32- 02=
and Area (continued) 2
= < F= 7 x A = 9
3 Evaluate the integral. 0
0
4 You could have just found the area of the triangle with a base
of 3 and a height of 6. Therefore, AR = 9.

A= 1 b h = 1 $ 3 $ 6 = 9
2 2

AREA OF A BOUNDED REGION: TRIGONOMETRIC FUNCTION y
Find the area of the region bounded by the graphs of y = cos x, y = 0, y = cosx
and x = 0.
R
1 Sketch a diagram; note that cos x = 0 when x = π2.
0
x=0 π y=0

2

2 Set up the appropriate integral representing the area of the AR= ͐ π/2 cos xdx
given region.
0

3 Evaluate the integral. = 6sin x @π/2 = sin b π l - sin ^0h = 1 - 0 = 1
0 2

206

10Introduction to the Integral chapter

Some Properties of the Definite Integral ͐5 _ln x + e x + sin x i dx = 0
5
1 If the function f is defined at x = a, then ͐a f ^xh dx = 0.
a
It simply states that the area from x = a to x = a is 0.
There is no work required; the area is just 0.

2 If the function f is integrable on [a,b], then ͐2 f ^xh dx = -͐5 f ^xh dx
5 2
͐b f ^xh dx = - ͐a f ^xh dx. This makes sense, since by
ab

switching the order of the limits of integration, you just

switch the order of the subtraction when you evaluate

the integral.

3 For a < b < c, if the function f is integrable on [a,b], [b,c], y

and [a,c], then ͐c f ^xh dx = ͐b f ^xh dx + ͐c f ^xh dx. y = f(x)
a ab

ab x
c

In terms of area,

͐1 4ca4f2^x4h4dx3 = ͐1 4ba4f2^x4h4dx3 + ͐1 4cb4f2^x4h4dx3
area of green area of red area of blue

region region region

207

The Definite Integral
and Area (continued)

4 If the function f is $͐2 ͐2 x4 2 12 b146 1 12 b145l =
integrable on [a,b] 1 1 4 4
12x 3 dx = 12 x3 dx = 12 < F = - l = 45

1

and k is a constant,

$ $then͐b ͐b
a k f ^xh dx = k a f ^xh dx.

5 If the functions f and g are integrable on [a,b], then ͐e b 1 + exl dx
1 x
͐b _ f ^xh ! g ^xhi dx = ͐b f ^xh dx ! ͐b g ^xh dx.
a aa = ͐e 1 dx + ͐e ex dx
1 x 1

= ln ^eh - ln ^1h + ee- e1

= 1 - 0 + ee- e1

= 1 + ee- e1

208

Second Fundamental

10Theorem of Calculus Introduction to the Integral chapter

Let f be a function that is continuous on [a,b], and let x be any number in [a,b]; then

d ͐ x f ^th dt = f ^xh .
dx a

Remember, if you let F ^ x h = ͐ x f ^th dt , then the theorem is just saying that F’(x) = f(x). For example,
a

d ͐x _ln t + eti dt = ln x\+ ex . Just replace f(t) with f(x).
dx 2 1 44 2 44 3 f ^xh

f ^th

EXAMPLE 1

Find d ͐x _t 2 - 2ti dt . d ͐x _ t 2i - 2t dt = x 2 -\2x or just x2- 2x
x 3 dx 3 \ f ^xh
f ^th

EXAMPLE 2

Find d ͐x _sin t + et- t3i dt = sin x + ex- x3
dx 5

or just sin x + ex- x3.

EXAMPLE 3

Find d ͐x t + ln t dt. ͐d x t + ln t dt = x + ln x or x + ln x
dx 2 t3 t3 x3 x3
3 dx 2 3

209

Second Fundamental Theorem
of Calculus (continued)

SECOND FUNDAMENTAL THEOREM OF CALCULUS: TOP LIMIT IS A FUNCTION OF X
What happens when the top limit of integration is some function of x, other than x itself?

u$Ifis a of x, d ͐u du .
function then dx a f ^th dt = f ^uh

1 Find d ͐ x3 eti dt . x3
dx 4 _ln t + $͐d

dx
4 _ln t + eti dt = 9ln _ x3i + ex3C 3Xx 2
1442443 144424443
f ^th der. of

f _x3i x3

$or just 8ln x3+ ex3B 3x2

d ͐ln x ln x͐ $d 1
dx π Xx
π _t2+ dx
2 Find sin ti dt . _t2+ sin ti dt = 9^ln xh2+ sin ^ln xhC der. of
1 44 2 44 3 14444244 443 ln x

f ^th f ^ln xh

210

11chapter

Techniques of
Integration

This chapter introduces many techniques of Power Rule: Simple and General . . . . 212
integration, the process of finding the integral
of a function. A lot of these techniques depend upon Integrals of Exponential Functions . . . 220
your being able to recognize a pattern in the way
the function is, or can be, written—such as in an Integrals That Result in a Natural
exponential function or some power of a function, Logarithmic Function . . . . . . . . . . . . 223
or a function whose integral will result in a
logarithmic function. Integrals of Trigonometric
Functions. . . . . . . . . . . . . . . . . . . . . . . 226
This chapter introduces integration techniques
such as integrals involving trigonometric functions, Integrals That Result in an Inverse
or integrals that result in inverse trigonometric Trigonometric Function. . . . . . . . . . . 232
functions. Sometimes multiple techniques are
required to integrate a given function. The use of Combinations of Functions and
algebraic substitution covers some integrals that do Techniques . . . . . . . . . . . . . . . . . . . . . 235
not seem to fit any other pattern of integration. The
chapter concludes with solving some differential Algebraic Substitution . . . . . . . . . . . . . 237
equations.
Solving Variables Separable
Differential Equations . . . . . . . . . . . . 240

Power Rule: Simple
and General

There are two versions of the Power Rule to consider when finding the integral of a function: the
Simple Power Rule, in which you integrate powers of the term x, and the General Power Rule, in
which you integrate powers of a function of x.

Simple Power Rule

͐x n dx = x n+1 + c, where n ! -1
n+1

EXAMPLE 1 ͐ ?xn

͐x2 dx x2 dx

1 Start with the given expression.

2 Integrate, applying the Simple Power Rule, by increasing the power of x to 3 @xn + 1
and then dividing the new term by 3. x3
= T3 + c

n+ 1

= x3 + c
3

EXAMPLE 2

͐ 1 dx
x2

1 Rewrite the integrand as a negative power of x. = ͐x- 2 dx

212

11Techniques of Integration chapter

2 Apply the Simple Power Rule by adding 1 to the exponent and dividing the Axn + 1
new term by the new exponent of –1. x- 1
= -T1

n+ 1

3 Simplify the resulting expression. = - 1 + c
x

EXAMPLE 3 ͐ 3 x2 dx
͐= x2/3 dx
͐ 3 x2 dx
= x 5/3 + c
1 Rewrite the radical term as an exponential term. 5

2 Apply the Simple Power Rule and simplify the result.

3

= 3 x 5/3+ c
5

3 Rewrite the result as a radical term since the original integrand was a = 33 x5 + c
radical term. 5

= 3 x 3 x2 + c
5

213

Power Rule: Simple and
General (continued)

EXAMPLE 4

͐ d 3x4- 4x 2 + 7 n dx
x2

1 Rewrite the original rational expression as three ͐ 3x 4 - 4x 2 + 7
separate rational expressions. x2
d n dx

= ͐ d 3x 4 - 4x 2 + 7 n dx
x2 x2 x2

2 Simplify each rational expression. = ͐ _3x2- 4 + 7x-2i dx

3 Break up the expression into three separate integrals. = 3͐x2 dx - ͐4dx + 7͐x-2 dx

Remember, you can just move the constants, 3 and 7,
outside of the integral sign.

4 Evaluate each integral by adding just one “+ c” at = 3 d x3 n - 4x + 7 d x-1 n + c
the end. 3 -1

= x 3- 4x - 7 + c
x

General Power Rule

If u is a function of another variable, say x, then ͐un du = un+ 1 + c, where n !- 1 .
n+
1

Remember that du is just the derivative of the function u. Another way to view this is

_orig. funct.i n+ 1
n+ 1
͐ _orig. n $ _ der. of orig. funct.i = +c

funct.i

214

EXAMPLE 1 11Techniques of Integration chapter

͐_x2+ 6 ͐ Hu ?n Ddu

5i 2xdx _ x2+ 5i 6 2xdx

1 Identify the u, n, and du for this problem.

n +1 _Hx 2+u 5i@n+7 1
= T7 + c
2 Apply the General Power Rule: _orig. funct.i
n+1 +c n +1

= 17_x2+ 5i + c

EXAMPLE 2

͐x2_ x3- 4

7i dx

1 Although this integral looks like it might be a Simple Power ͐x2_ x3- 4
Rule situation, let’s move the terms around to be sure.
7i dx

͐ _ x3- 4
$= x2 dx
7i

2 Unfortunately, the x2dx term is not quite the derivative of the 1$=_ x 3- 4 3x[2 dx
inside function x3 – 7. Since the derivative of x3 – 7 = 3x2dx, you 3 \ du
7i
can multiply by 3 inside the integral, and compensate for it by
multiplying the outside by 13. u

215

Power Rule: Simple and R 5 V
General (continued) S_x3- W
= 1 S 7i W+ c
3 Now that the integral fits the General Power Rule pattern, you can 3 S W
just increase the exponent 4 by 1 to a 5, and then divide this term 5
by 5.

TX
5
_x3-
7i
= 15 + c

EXAMPLE 3 (USING CHANGE OF VARIABLE OR U-SUBSTITUTION TECHNIQUE)

͐x2_ x3- 4

7i dx

1 After writing the original integrand in a more useful form (it looks ͐x 2_ x 3- 4 dx
like a General Power Rule pattern), you have the expression at right.
7i

͐ 4

7i
$= _ x 3- x2 dx

2 Let’s try a different approach by using let u = x3- 7, then du = 3x2 dx
what is called the “change or variable” or
“u-substitution” technique, letting u = the or 1 du = x2 dx
inside function and then proceed as shown 3
at right.
Notice that you now have all the terms of
the original integrand written in terms of
the new variable u.

3 Now substitute the u terms found in Step 3 for the corresponding parts in = ͐u 4 $ 1 du
Step 2. 3

216

11Techniques of Integration chapter

4 Move the 1 out in front of the integral and apply the General Power Rule. = 1 ͐u 4 du
3 3

= 1 $ u5 + c
3 5

= u5 + c
15

5 You need to return to a function in terms of x, _x3- 5 _x3- 5
not u. So substitute for u = x3 – 7 in the result
7i 7i
from Step 4. = 15 + c or just 15 + c

EXAMPLE 4 (WITH LIMITS OF INTEGRATION) ͐2 5x2 dx

͐2 5x2 dx 0 x3+ 1

0 x3+ 1 ͐= 2 5x 2 dx
1 Rewrite the integrand by bringing the radical term from the 0
_x3+ 1/2
denominator up to the numerator as an exponential term
instead. 1i

2 Move some of the terms so that it looks more like a General ͐= 2 - 1/2
Power Rule situation.
0 5x 2_ x3+ 1i dx

2 - 1/2

0 _ x 3+ 1i
͐ $= 5x2 dx

217

Power Rule: Simple and 2 - 1/2
General (continued)
0 _ x 3+ 1i
3 Rewrite the term 5x2 as the derivative of the inside $ $͐= 5 x2 dx
function x3 + 1 (that is, you need a 3x2, not a 5x2).
1 2 - 1/2
4 You are finally ready to apply the General Power Rule. 3
0 _ x 3+ 1i
$ $ $͐= 5 3x2 dx

5 2 - 1/2
3 0
1i
$ $͐=
_ x3+ 3x[2 dx
1 44 2 44 3 du
u

5 R x 3 + 1/2 V2
3 S_ W
= S 1 1i W
SS 2 WW
T X0
10 2
= 3 8
x3+ 1B
0

5 Plug in the limits of integration to simplify the result. = 10 8 23+ 1- 03+ 1B
3

= 10 ` 9- 1j
3

= 10 ^2h = 20
3 3

EXAMPLE 5

0͐ π/2 sin x dx

cos x

1 First, note that the derivative of sin x is cos x. Use this fact to ͐ π/2 sin x dx
set up the integrand in the General Power Rule format.
0 cos x

͐= π/2 Hu Hdu
^sin xh1/2 cos xdx
0

218

11Techniques of Integration chapter

2 Determine the integral and transfer the limits of R Vπ/2
integration. SSTSS ^sin23x h3/2 W
= W
WW
X0
π/2
= 2 :
3 ^sin xh3D

0

3 Evaluate the last expression by plugging in the limits of R 3 V
integration. 2S π W
= 3 SS bsin 2 l - ^sin 0h3 WW

T X
= 238 13 - 03B

= 2
3

219

Integrals of Exponential
Functions

If u is a function of some other variable, say x, then ͐eu du = eu+ c and ͐au du = a u + c . Another
ln
a

$way to write the first integral above is ͐esome funct. ^der. of funct.h = ethat +funct. c . For example,
͐ $ B?u du ?u
e3x 3dx = e3x+ c or just e3x+ c .

EXPONENTIAL INTEGRAL: EXAMPLE 1 ͐xex2 dx

͐xex2 dx = ͐ex2 xdx

1 Rewrite the integrand in a form that is closer to the
exponential integral pattern.

1 $= 1 ͐ eTx2 2xYdx
2 2
Notice how the and the 2 were used to get the correct eu du

du term.

2 Use the exponential integral pattern to finish the problem. = 1 e x2 + c
2

EXPONENTIAL INTEGRAL: EXAMPLE 2

͐ eln x

x dx

1 Rewrite the integrand to try to make the derivative of the exponent, lnx, ͐ eln x

eln x. d 1 x dx
dx x
follow the term Remember that ^ln xh = . Ddu

Bu 1
x
eln x
͐ $= dx

220

2 Apply the exponential integral formula. 11Techniques of Integration chapter

= eln x + c

EXPONENTIAL INTEGRAL: EXAMPLE 3 (WITH LIMITS OF INTEGRATION) ͐9 e x dx

͐9 e x dx 1x

1x

1 Rewrite the integrand so it follows the

͐ $esome funct. ^der. of funct.h = ethat +funct. c pattern.

͐= 9 x1/2

ex dx1 1/2

͐ $9

= e x dxx1/2 - 1/2
1

2 You don’t quite have the correct derivative of the exponent yet, G9 Au du
but by inserting a 1⁄2 inside and a corresponding 2 outside, 1x1/2 - 1/2
you’ll get what you need.
e 2 x dx1
3 Complete the formula for the exponential integral and carry $ ͐ $= 2
over the limits of integration.
= 1 7e x 9
4 Plug in the limits and simplify the result. 2
A
1

= 1 8e 9- eg1 B
2

= 1 7 e 3- eA
2

221

Integrals of Exponential
Functions (continued)

EXPONENTIAL INTEGRAL—BASE OTHER THAN e: EXAMPLE 1

͐25x+ 7 dx

1 You are trying to fit this into the ͐au du = au + c formula. To do ͐25x+ 7 dx
ln a

so, you need to have the derivative of the exponent (5x + 7) follow 1 Gu Bdu
5 5dx
25x + 7
the exponential term. $ ͐ $=

2 Apply the formula for the non e base exponential = 15< 25x +7 F + c
integral. ln 2

= 2 5x + 7 + c or 25x +7 +c or 25x +7 + c
5 ln 25 ln 32
ln 2

EXPONENTIAL INTEGRAL—BASE OTHER THAN e: EXAMPLE 2 $͐x 5x2 dx

$͐x 5x2 dx

1 Try to fit the original integrand into the exponential integral pattern.

$͐= 5xx2 xdx

$ $͐=1 ?u Ddu
2 2xdx
5x2

2 Complete the integral using the exponential = 1 R ?u V
integral formula. 2 S W
S 5x2 W+
S l\n 5 W c
T ln X
of base

= 5x2 5 + c or 5x2 +c or 5 x2 + c
2 ln ln 52 ln
25

222

Integrals That Result in a 11chapter
Natural Logarithmic Function

If u is a function of some other variable, say x, then ͐ du = ln u + c. Another way to write this is
u

Gdu
3x 2 3x2 dx
͐ der. of funct. = ln funct. + c. For example, ͐ x3- 5 dx = ͐ x3[- 5 = ln x3- 5 + c.
funct. 14424 3

u ln u

NATURAL LOG INTEGRAL: EXAMPLE 1

͐ x +1 1 dx
x2+ 2x -

1 You want the top to be the derivative of the bottom. Since ͐ x +1 dx
x2+ 2x -
d _ x 2 + 2x - 7i = 2x + 2, you just need to multiply the 1
dx
1 = 1 $ ͐ 2 ^x + 1h dx
numerator by 2 and then compensate with a 2 outside the 2 x2+ 2x - 1

integral symbol. 6 44 7du 44 8
^2x + 2h dx
= 1 $ ͐ 1x42 +4 22x4-4 13
2

u

2 Complete the formula, which results in ln u + c . 1 6 447u 44 8
2 x2+ 2x - 1
= ln +c

or just 1 ln x2+ 2x - 1 +c
2

1/2

or ln a x2+ 2x - 1 k + c

or ln x2+ 2x - 1+ c

223

Integrals That Result in a Natural
Logarithmic Function (continued)

NATURAL LOG INTEGRAL: EXAMPLE 2 (USING A U-SUBSTITUTION)

͐ e2x + 1 dx
e2x+ 2x

1 Since you are trying to make this fit the ͐ du let u = e2x+ 2x, then du = _2e2x+ 2i dx
u

pattern, let u = e2x + 2x and then proceed as du = 2 _e2x+ 1i dx

shown at right. or 1 du = _e2x+ 1i dx
2

2 Substitute the terms found in Step 1 into the appropriate spots in the ͐ e2x+ 1 dx
original integrand. e2x+ 2x

Note: All expressions containing the variable x have been replaced with = ͐ _ e2x + 1i dx
u variable terms. e2x + 2x

͐ b 1 l du
2 u
=

= 1 $ ͐ du
2 u

3 Now that the integral is in the ͐ duu, complete the formula with the = 1 8ln u B+ c
2
ln u + c portion.
= 1 ln e2x + 2x + c
4 You need to get back to a solution in terms of x, not u, so 2
substitute u = e2x + 2x.

or just 1 ln e2x + 2x + c
2

224

11Techniques of Integration chapter

NATURAL LOG INTEGRAL: EXAMPLE 3 (USING A U-SUBSTITUTION)

͐3 x x3 4 dx
2 4+

1 Since you are trying to fit this to a natural log form let u = x4+ 4, so that du = 4x3 dx

͐ du let u be the denominator, so u = x4 + 4. or 1 du = x3 dx
u 4

2 Since you are making a u-substitution, you may as well u = x4 + 4
change the original x limits to the new u limits of for x = 3 → u = 34 + 4 = 85
integration. for x = 2 → u = 24 + 4 = 20

3 Make a lot of substitutions, including ͐3 x x3 4 dx
the limits of integration, so that the 2 4+
original problem changes from x
terms and limits to u terms and limits. ͐= 3 x3 dx % the lim . of int . are x values
2 x4+ 4

͐85 b 1 l du
4
= 20 % the lim . of int . are u values
u

$ ͐=1 85 du
4 20 u

4 Complete the integral, carry over the limits of = 1 8ln 85
integration, plug them in, and simplify the result. 4
uB
TIP 20

By changing from x limits to u limits of integration, you = 1 6ln 85 - ln 20@
don’t have to change back to x terms at the end. 4

= 1 ln b 85 l
4 20

or other forms, such as

1 ln b147 l or ln b 17 1/4 or ln 4 17
4 14 4
l

225

Integrals of Trigonometric
Functions

If u is a function of some ͐ cos udu = sin u + c ͐ sec u tan udu = sec u + c
variable, say x, then: ͐ sin udu = -cos u + c ͐ csc2 udu = -cot u + c
͐ sec2 udu = tan u + c ͐ csc u cot udu = -csc u + c

The integral formulas written For example, since
above are just the result of reading
backward the derivative formulas d ^sin uh = cos udu " cos udu = sin u + c
for the six trigonometric functions. dx

Similarly, since

d ^sec uh = sec u tan udu " ͐ sec u tan udu = sec u + c
dx

If u is a function of some variable, ͐ tan udu = - ln cos u + c or ln sec u + c
say x, then: ͐ cot udu = ln sin u + c
͐ sec udu = ln sec u + tan u + c
͐ csc udu = ln csc u - cot u + c

The integral formulas above are a bit more der. of
difficult to verify. One example is shown at right.
Fthe funct.
cos u
͐ cot udu = ͐ si[n u du = ln si[n u +c

the funct. the funct.

226

11Techniques of Integration chapter

TRIGONOMETRIC INTEGRAL: EXAMPLE 1

͐ cos 5xdx

1 To make this fit the ͐ cos udu pattern, you need the derivative ͐ cos 5xdx

of 5x, namely 5, to follow the cos5x term. So insert a 5 and 1 $ ͐ $
5
compensate with a 1 outside the integral symbol. = fcos 5Tx p 5Vdx
5
u du

2 You can now complete the integral using the right-hand side of the = 1 sin 5Tx + c
5
formula ͐ cos udu = sin u + c . u

TRIGONOMETRIC INTEGRAL: EXAMPLE 2 ͐x2 sin _x3i dx

͐x2 sin _x3i dx $= ͐ sin _x3i x2 dx

1 Let’s try the u-substitution method on this integral. But first,

rewrite the integrand so it looks more like the ͐ sin udu pattern.

2 Let u = x3 and then find du. let u = x3, then du = 3x2 dx

or 1 du = x2 dx
3

227

Integrals of Trigonometric $= ͐ sin _x3i x2 dx
Functions (continued)
= ͐ sin u $ 1 du
3 Substitute the expressions from Step 2 for the appropriate terms in 3
Step 1.
= 1 $ ͐ sin udu
4 You can now complete the ͐ sin udu = - cos u + c formula and 3

replace the u with x2 to finish the problem. = 13^-cos uh + c
= -13 cos _x3i + c
TRIGONOMETRIC INTEGRAL: EXAMPLE 3
͐ sec 3x tan 3xdx
͐ sec 3x tan 3xdx
1 This integral looks like the ͐ sec u tan udu = sec u+ c form; = 1 ͐ sec 3Tx tan 3Tx $ 3Vdx
3
you need a 3 to be the derivative of 3x.

u u du

2 Complete the right-hand side of the red formula = 1 sec 3Tx + c or just 1 sec 3x + c
above. 3 3
u

228

11Techniques of Integration chapter

TRIGONOMETRIC INTEGRAL: EXAMPLE 4

͐ π/2 1 sin x x dx
0 + cos

1 This looks like a natural log integral form, with u = 1 + cos x :
because the top is almost the derivative of the
bottom. Let u = 1 + cosx and then find du. At for x = π " u = 1 + cos b π l = 1 + 0 = 1
the same time, let’s change from the given x 2 2
limits of integration to the new u limits.
for x = 0 " u = 1 + cos 0 = 1 + 1 = 2

2 Replace all x terms and limits with ͐ π/2 1 sin x x dx
their appropriate u term counterparts. 0 + cos

= ͐ π/2 sin xdx %x terms and x lim. of int.
0 1 + cos x

= ͐1 - du %u terms and u lim. of int.
2 u

͐1 du
u
=- 2

3 Notice that the upper limit of integration is smaller than the lower = - ͐2 du F
limit. Since you are probably used to having the upper limit bigger u
than the lower limit, switch the limits and also take the opposite of the <- 1
integral.
= ͐2 du
1 u

4 Complete the natural log integral form with the right-hand side of the 2

formula ͐ du = ln u. (The “+ c” was dropped because there are limits = 8ln u B
u 1

of integration involved in this problem.) = ln 2 - ln 1
= ln 2 - ln 1
5 Plug in the limits and simplify the result. = ln 2 - 0
= ln 2

229

Integrals of Trigonometric
Functions (continued)

TRIGONOMETRIC INTEGRAL: EXAMPLE 5

͐ cos4 x dx
csc x

1 In searching for an appropriate integration technique to use here, ͐ cos4 x dx
the one that first comes to mind is a natural log. But in this case, csc x
the top is not the derivative of the bottom, so the natural log
won’t work. Let’s try rewriting the integrand to see if some other = ͐ cos 4 x $ 1 x dx
technique presents itself. csc

= ͐ cos4 x $ sin xdx

= ͐ ^cos xh4$ sin xdx

2 Now it looks like a General Power Rule pattern, but the $= -͐ ^Hcous xh?n4 6 44 7du 44 8
derivative of the inside function, cos x, is actually –sin x, ^-sin xh dx
so we need a negative sign to get the General Power Rule
just right.

3 Complete the General Power Rule formula: ͐un du = u n +1 + c Hu @n+1
n+1 = -^cosT5xh 5 + c

n +1

= -cos5 x + c

230

11Techniques of Integration chapter

TRIGONOMETRIC INTEGRAL: EXAMPLE 6 ͐ sec4 x tan xdx

͐ sec4 x tan xdx = ͐ ^sec xh4$ tan xdx

1 Let’s use a bit of an unusual strategy on this integral. First,
rewrite the integrand, making the power of sec x more
obvious.

2 This almost looks like a General Power Rule pattern, except = ͐ ^sec xh3$ sec x tan xdx
= ͐ ^se\c xh3 $ s1e4c4x2tan4x4dx3
that the derivative of sec x is sec x tan x. So “borrow” a sec x
term from the (sec x)4. u du

3 Now that your integral fits the ͐un du = un+ 1 + c pattern, Hu @n+ 1
n+ ^sec xh 4
1 T4

complete the right-hand side of the General Power Rule = +c

formula. n+ 1

= 1 sec 4 x + c
4

231

Integrals That Result in an
Inverse Trigonometric Function

If u is a function of some variable, say x, and a is some constant, then ͐ du = arcsin u + c
a2- a
u2

͐ du = 1 arctan u + c
a2+ u2 a a

͐ du = 1 arc sec u +c
u2- a2 a a
u

At right are brief Bdu ?u
examples of each ͐ 3dx ͐ 3dx 3x 3x
of these integral 25 - 9x2 = = arcsin S5 + c or arcsin 5 + c
formulas. S52 ^3Yxh2
a - u a

Fdu Au 5x 2
10xdx 10xdx 1 5x 2 1 7
͐ 49 + 25x 4 = ͐ = 7 arctan S7 +c or 7 arctan +c
S72 + _5x 2i2
a [ a

u

͐ 5dx 16 = ͐ Bdu S42 = 1 arc sec Au + c = 1 arc sec 5x +c
25x 2 - 5dx a S4 5x 4 4
5x 5Tx ^5Yxh2 - S4
a
u a

u

INVERSE TRIGONOMETRIC INTEGRAL: EXAMPLE 1 ͐ dx

͐ dx 25 - 4x2

25 - 4x2 = ͐ dx ^2xh2
S52 - Y
1 How do you determine which inverse trigonometric form to use? In a
this case, the denominator has a radical that contains a constant u
minus a function square, as in a2- u2 . So this is an arcsin form.
You need to write the integrand in that form.

232

11Techniques of Integration chapter

2 You’re close to the correct form, except that the numerator is not Bdu
2dx
the derivative of the u term, 2x. Since d ^2xh = 2, you need to = 1 ͐
dx 2 S52 - ^2xh2
a Y
insert a 2 on top of the integrand and compensate with a 1⁄2
u

outside.

3 Now the integral fits the left side of the formula R ?u V
S 2x W
͐ du = arcsin ua+ c , so just complete the = 1 Sarcsin S5 W+ c or 1 arcsin 2x + c
a2- 2 SS WW 2 5
u2 a X
T
right-hand side.

INVERSE TRIGONOMETRIC INTEGRAL: EXAMPLE 2

͐ x 2 dx 6
16 +x

1 If this is an inverse trigonometric integral, it has to be arctan, because ͐ x 2 dx 6
there is no radical in the denominator. Rewrite the integrand to try to get 16 +x
it into the arctan pattern.
= ͐ x2 dx
42+ _x3i2

2 Let’s use the u-substitution this time—it appears that let u = x2 , then du = 3x2 dx
u = x3. Find du also.
du = x2 dx
or 3

233

Integrals That Result in an Inverse = ͐ x2 dx
Trigonometric Function (continued)
42+ _x3i2
3 Substitute the u terms from Step 2 for the appropriate x terms in
Step 1.

du

= ͐ 3 u2
42+

= 1 ͐ du
3 42+ u2

4 With a = 4, this now fits the arctan formula: = 13; 1 arctan u E + c
4 4
͐ du = 1 arctan u + c. Now complete the right-hand side of
a2+ u2 a a 1 u
12 4
the formula. = arctan + c

5 Last, make the substitution of u = x3 to = 1 arctan x3 + c or 1 arctan x3 + c
complete the problem by returning to x terms. 12 4 12 4

234

Combinations of Functions 11chapter
and Techniques

You will occasionally encounter an integral that by itself cannot be integrated. In some of these
cases you will have to first alter the form of the integrand in order to use multiple techniques to
complete the integration process.

“Combo” Technique: General Power Rule and an Arcsin ͐ x + 4 dx

͐ x + 4 dx 9 - x2

9 - x2 J x+ 4 N
1 This does not fit any of the forms you have studied 9 - x2 9-
= ͐ KK
so far. Split the original integral into two separate
integrals to see if that helps. L

x 2 OOdx
P

= ͐ x dx + ͐ 4 dx
9 - x2 9 - x2

2 The blue integral is in a General Power Rule $ $͐= _9 - x2i- 1/2 xdx+ 4 ͐ dx
form, and the red integral is in an arcsin form. 32- ^xh2
Modifying the integrands further will make
these forms more apparent.

1 ͐_ -1/2 ͐ dx
2 32 - ^ xh2
i
$ $ $=- 2
3 The first integral needs a –2 inside 9 - x ^-2h xdx + 4

(with a - 1 outside); the second
2

integral’s form is fine.

4 Use the General Power Rule for the blue and an 1 i2 1/2 x
arcsin for the red integral. $= - 2 - D + 3 +
:_9 x 4 ;arcsin E c

= - 1 9 - x 2 + 4 arcsin x + c
2 3

235

Combinations of Functions
and Techniques (continued)

Regarding “Look-Alike” Integrals

The integral forms in each ͐ 9 x3 dx and ͐ 9 x x4 dx ͐ x3 x4 dx and ͐ 9 x x4 dx
colored pair listed at right + x4 + 9+ -
are frequently mistaken for
one another.

1 Take a closer look at the red pair as you rewrite ͐ x 3 dx ͐ 9 x x 4 dx
their integrands to reveal the particular technique + +
appropriate to that integral. 9 x 4

1 ͐ 4x 3 = ͐ x dx
4 9\+
$= x 4 9 + _ x 2 2

i

natural log $= 1 ͐ 2x dx
2 + _x
9 2i2
1 44 2 44 3
arctan

2 Take a closer look at the blue pair ͐ x3 ͐ x x4 dx
as you rewrite their integrands to 9-
reveal the particular technique 9 + x 4 dx
appropriate to that integral. =͐ x x 2i2 dx
͐= _9 + x 4i- 1/2 x3 dx -_

1 ͐ _9 + x 4i- 1/2 4x3 dx 9
4
14444424444 3
$ $= $=1 ͐ 2x dx
2 9 - _x2i2
general power rule

144424443
arcsin

236