8.3. INTEGRATION BY PARTS 195
8.3 Integration by PartsAnother technique for finding antiderivatives is called integration by parts and is based onthe product rule. Recall the product rule. If u′ and v′ exist, then
(uv)′ (x) = u′ (x)v(x)+u(x)v′ (x) . (8.2)
Therefore,(uv)′ (x)−u′ (x)v(x) = u(x)v′ (x)
Proposition 8.3.1 Let u and v be differentiable functions for which∫u(x)v′ (x) dx,
∫u′ (x)v(x) dx
are nonempty. Then
uv−∫
u′ (x)v(x) dx =∫
u(x)v′ (x) dx. (8.3)
Proof: Let F ∈∫
u′ (x)v(x) dx. Then
(uv−F)′ = (uv)′−F ′ = (uv)′−u′v = uv′
by the chain rule. Therefore every function from the left in 8.3 is a function found in theright side of 8.3. Now let G ∈
∫u(x)v′ (x) dx. Then (uv−G)′ =−uv′+(uv)′ = u′v by the
product rule. It follows that uv−G ∈∫
u′ (x)v(x) dx and so G ∈ uv−∫
u′ (x)v(x) dx. Thusevery function from the right in 8.3 is a function from the left.
Example 8.3.2 Find∫
xsin(x) dx.
Let u(x) = x and v′ (x) = sin(x). Then applying 8.3,∫xsin(x) dx = (−cos(x))x−
∫(−cos(x))dx =−xcos(x)+ sin(x)+C.
Example 8.3.3 Find∫
x ln(x) dx.
Let u(x) = ln(x) and v′ (x) = x. Then from 8.3,∫x ln(x) dx =
x2
2ln(x)−
∫ x2
2
(1x
)=
x2
2ln(x)−
∫ x2=
x2
2ln(x)− 1
4x2 +C
The next example uses a trick.
Example 8.3.4 Find∫
arctan(x) dx.
Let u(x) = arctan(x) and v′ (x) = 1. Then from 8.3,∫arctan(x) dx = xarctan(x)−
∫x(
11+ x2
)dx