Lecture 16

Calculating Antiderivatives

Tables of Derivatives and Antiderivatives

f f '

xn

n x( )n 1

( )ln x1

x

ex

ex

f f '

un

n u( )n 1

du

( )ln udu

u

eu

eu

du

With Chain RuleBasic Derivatives

f d f u

n u( )n 1

du un

du

u( )ln u

eu

du eu

As Table of Integrals

d

u

nu u

( )n 1

n 1C d

1

uu ( )ln u C

d

eu

u eu

C

Basic Integrals

General Rules

d ( )f x ( )g x x d

( )f x x d

( )g x x

d c ( )f x x c d

( )f x x

d

3 x

45 x2

2 x 7 x

3 d

x4

x 5 d

x2

x 2 d x x 7 d

1 x

3 x5

5

5 x3

3

2 x2

27 x C

=

=

d

4 x3

x5 e

xx 4 d

x x 3 d

1

xx 5 d

ex

x

= 4 x

3

2

3

2

3 ( )ln x 5 ex

C

d

5 x2

x3

xx d

x5 x

3

2x

5

2x

=

=

d

1

xx 5 d

x

3

2x d

x

5

2x

= 2 x 2 x

5

2 2

7x

7

2C

=

x

1

2

1

2

5 x

5

2

5

2

x

7

2

7

2

C

d

dx( )ln 1 x

11 x

d

11 x

x ( )ln 1 x

d

x3

5 x2

x 21 x

x

so

x3

5 x2

x 2 ( ) x2

6 x 7 ( )x 1 9

x3

5 x2

x 2x 1

x2

6 x 79

x 1

d

x3

5 x2

x 2x 1

x d

x2

6 x 7 x 9 d

1x 1

x

1

3x3

3 x2

7 x 9 ( )ln 1 x C=

Integrating by Substitution

d

u

nu u

( )n 1

n 1C

d

( )1 x

353 x2

x

u 1 x3

du 3 x2

dx

n = 5

= d

u5

u

= 1

6u6

C

= 1

6( )1 x

36

C

Additional Example

d

e( )x2

x x

Looks like it might be of form d

eu

u If so then u x2

d

eu

x xdu 2 x dx du

2x dx

d

1

2e

uu = 1

2d

eu

u 1

2e

uC 1

2e( )x2

C=

Calculating by substitution

• Match (at least partially) to a form that is known

• Guess “u”• Calculate du• Replace all occurrences of x and dx by u

and du• Only adjustments with constants are

permitted• Doesn’t always work

d ( )f x x