Question 1: How do we find the antiderivative of functions involving products?
Let’s look at the Product Rule for Derivatives. For two functions u and v, the derivative is
d
uv   vu  uv dx
The strategy for integration by parts is discovered by taking the antiderivative of each of
these terms,

d
uv  dx   vu dx   uv dx dx
On the left side, the antiderivative of a derivative results in uv . This leaves us with
uv   vu dx   uv dx By solving for the second term on the right, we get the integration by parts strategy,
 uv dx  uv   vu dx We could solve for the first terms on the right and find a similar rule. Either rule is
acceptable, but we’ll use the rule above to carry out antiderivatives.
Integration by Parts Rule
If u and v are differentiable functions, then
 uv dx  uv   vu dx
The next several examples demonstrate how to apply this rule.
Example 1
Evaluate the Indefinite Integral
Evaluate

xe x dx .
2
Solution To apply the integration by parts rule, we need to match the
integrand xe x with the integrand on the left side of the rule. There are
two possible choices: u  x and v  e x or u  e x and v  x . Let’s try the
first choice.
To apply the rule, we also need to know u  and v . We find these by
taking the appropriate derivative and antiderivative:
Derivative
u  x 
 u  1
Antiderivative
 v  ex
v  e x 
Now substitute these pieces into the integration by parts rule:

x
x
x
x e dx  x e   e  1 dx
u v
u v
v
u
The integration by parts results in a new indefinite integral. However,
the new indefinite integral is much easier to evaluate. Since the
antiderivative of e x is e x  C , we can write down the antiderivative as

xe x dx  xe x  e x  C
Technically, we should be subtracting C. Since C is an arbitrary
constant, adding or subtracting C does not affect the antiderivative.
Let’s abbreviate these steps and write them down in a compact form.

x
x
x
x e dx  x e   e  1 dx
u v
u v
u
v
 xe  e  C
x
x
Derivative
u  x 
 u  1
Antiderivative
 v  ex
v  e x 
The second choice for u and v leads to a new indefinite integral that is
difficult.
 e x dx  e    
x
u v
x
x2
2
u
v
x
 e x dx
2 
2
v
u
Derivative
u  e x 
 u  e x
Antiderivative
v  x 
 v
x2
2
3
The new integral is even more difficult than the original integral. For this
reason, it is better to choose u  x and v  e x .
When choosing u and v , we must make sure the new integral is one we can evaluate.
Since we need to find v from v , we also need to make sure the choice for v is one we
can find the antiderivative of.
Example 2
Evaluate the Indefinite Integral
Evaluate

x ln  x  dx .
Solution We need to identify the factors in the integrand with the factors
uv in the integration by parts rule. Let’s start by selecting u  x and
v  ln  x  . For this to be useful, we need to be able to take the
derivative of x to get u  and the antiderivative of v  ln  x  to get v. The
first task is easy, but the antiderivative of ln  x  is not one of the basic
function we know how to take the antiderivative of.
Let’s try choosing u  ln  x  and v  x . With this choice, we are able to
carry out the derivatives and antiderivatives:
1
x
x2
Antiderivative

 v
2
Derivative
 u 
u  ln  x  
v  x
Now we may make these substitutions into the integration by parts rule,
 x  x dx  ln
 x 
 ln
 
x2
2
u
v
u
v
x
 1 dx
2 x
2
v
u
 ln  x 
x2
x
  dx
2
2
Simplify integrand
 ln  x 
x2 x2
 C
2 4
Apply Constant Times a
Function Rule followed by
Power Rule
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This choice for u and v makes it possible to take the antiderivative of
v . It also makes the new indefinite integral much easier to compute.
In some cases, the integration by parts rule may need to be applied more than once.
Each time the new indefinite integral becomes simpler and eventually it is trivial to
solve.
Example 3
Evaluate the Indefinite Integral
Evaluate
x
2
 x  e x dx .
Solution Choose u and v and apply the integration by parts rule:

x
x 2  x  e x dx   x 2  x  e x   e
2 x  1 dx









v
v
u
u
v
u
Derivative
u  x 2  x 
 u  2 x  1
v  e x
Antiderivative

 v  ex
The new indefinite integral may also be solved using integration by
parts:

e x  2 x  1 dx   2 x  1 
e x   ex  2 dx







v u
v
u
u
v
  2 x  1 e x  2e x  C
  2 x  1 e x  C
Derivative
u  2 x  1 
 u  2
v  e x
Antiderivative

 v  ex
x
Simplify by factoring e from each
term and combining the remaining
terms
Put this in place of the indefinite integral in the first integration by parts
to get
 x
2
 x  e x dx   x 2  x  e x   2 x  1 e x  C
 e x  x 2  x  2 x  1  C
Factor e from each term and remove
parentheses
 e x  x 2  x  1  C
Combine like terms in parentheses
x
5
As before, we could pay careful attention to the sign of the arbitrary
constant C. Since it is arbitrary, simply bring it along in each step of the
calculation.
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