MAT1221 Handout 7.1
Product Rule – Reverse
d
 f ( x) g ( x)  f ( x) g ( x)  f ( x) g ( x)
dx
  f ( x) g ( x)  f ( x) g ( x) dx  f ( x) g ( x)
 f ( x) g ( x)dx   f ( x) g ( x)dx  f ( x) g ( x)
 f ( x) g ( x)dx  f ( x) g ( x)   f ( x) g ( x)dx
Let u  f ( x), v  g ( x) .
Then,
Integration By Parts
 f ( x) g ( x)dx  f ( x) g ( x)   f ( x) g ( x)dx
 f ( x) g ( x)dx   f ( x) g ( x)   f ( x) g ( x)dx
udv

uv

vdu


b
b
b
a
a
Example 1
a
Expectations
Supporting steps should be done on
the right hand column.
 xe dx
x
 xe dx
Let

Then,
x
u
, dv 
du

dx
, v
du 

1
Remarks 1. In the step of finding v, we always omit the arbitrary constant. Such a
constant does not affect the final form of the answer.
 xe dx
x
 x(e x  k )   (e x  k )dx
v   e x dx
 ex  k

Remarks 2. If we make a different choice such as…
 xe dx
x
Let

Example 2
 x ln xdx
u
, dv 
Then,
du

dx
, v
du 

 x ln xdx
Let
u
, dv 
du

dx
, v
du 

Then,
2
2
Example 3  ln xdx
1
2
 ln xdx
u
, dv 
du

dx
, v
du 

Let
1
Then,

Formula
d  1 kx  1 kx
kx
 1 kx
 e   e  kx   e  k   e
dx  k  k
k
e
kx
dx 
1 kx
e C
k
1
Example 4
 xe
2x
dx
0
1
 xe
2x
dx
Let
u
, dv 
du

dx
, v
du 

0
Then,

3
1
Example 5
x
x  1dx
0
1
x
x  1dx
Let
u
, dv 
du

dx
, v
du 

0
Then,

Classwork
Some integrals require IBP two times. The following is an example.
1
1. Evaluate
x e
2 3x
dx
0
1
(a) Show that after one IBP,
1
1 3 2
2 3x
3x
0 x e dx  3 e  3 0 xe dx
1
x e
2 3x
dx
Let
u
, dv 
du

dx
, v
du 

0
Then,

4
1
(b) Evaluate
 xe
3x
dx . (We plan to plug this into our answer in (a).)
0
1
 xe
3x
dx
Let
u
, dv 
du

dx
, v
du 

0
Then,

(c) Substitute your answer from (b) into the answer from (a) to get the final answer.
1
1
1 3 2
3x
0 x e dx  3 e  3 0 xe dx
2 3x
1
2
 e3  
3
3




5