Integration by Parts

MAT 272 Integration by Parts (IBP)
Spring 2019



Integration by Parts
The Antiderivative of a Product
We have an abundance of rules which are extremely particular; the integrand has to be exactly
the right hand-side of one of our rules. In particular, we cannot in general integrate a product of
functions, even if we know the antiderivatives of both factors individual.

2 2
x dx sin( x) dx x sin( x) dx
For example, we can certainly evaluate  and  , but  is not

2 ? x3
immediate. You might try
x sin( x) dx   3
cos( x) C
, but differentiating this supposed
antiderivative will not yield the original integrand. This is a naïve error that ignores the fact that there is
a Product Rule for differentiation because it seems convenient. The bottom line is that the antiderivative
of a product is not just the product of the antiderivatives of the two factors.

Whereas Substitution can be used to completely undo the Chain Rule (i.e., to antidifferentiate
products of exactly the form f ( g ( x )) g ( x ) ), there is no method for integrating a product in general
(i.e., to antidifferentiate any product f g ). Integration by Parts is as close as we can get to a general
method for products. Integrating a product is also different from undoing a Product Rule.


The problem at hand is f ( x) g ( x) dx , not f ( x) g ( x)  f ( x) g ( x) dx . Encountering product
functions f g is common, but finding an integrand which is the output of a Product Rule is so rare that
I get excited when it happens.

Integration by Parts

Suppose we wish to evaluate f ( x) g ( x) dx
. We start by assuming it occurred as a term in a
product rule. Then we work backwards to discover the original product which was differentiated and
subtract an appropriate integral from it.

d
 F ( x) G( x)  F ( x) G( x)  F ( x) G ( x)
Recall: The Product Rule is dx .

To make things easier, we leave out the function names ( f , g , F , G ) and instead use u and v .

(uv) uv  uv

Subtracting uv from both sides yields

(uv) uv uv


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, MAT 272 Integration by Parts (IBP)
Spring 2019

We then integrate both sides (with respect to the common independent variable of u and v , x )


(uv)dx  uv dx  uvdx
uv  v du  udv
Note that I’m using the differential relations du udx , and dv vdx .

Finally we are ready for IBP, which like Substitution may require trial and error, but unlike
substitution produces a new integral which may or may not be easier than the original!

Fact
To attempt to evaluate the integral of a product, call one factor u , and call the other factor (with the
original differential) dv . After calculating du and v , assemble the following integral equation:

u dv uv  v du
This integral equation (read left to right) writes the original integral in terms of another; the aim is take
an intractable integral and turn it into a simpler or more manageable one.

Example
x
Evaluate x e dx
.

First, we choose one of the two factors to be u . We can save time by choosing the factor that
gets simpler upon differentiating. Since differentiating the exponential yields yet another exponential,
but differentiating a polynomial decreases the degree by 1, in this case we choose u  x and call the
x
rest of the integrand dv e dx . It’s sometimes helpful to organize these parts in a table.

u  x  chosen Need to determine v .
dv e x dx  remainder
Need to determine du .
Then, compute the two missing pieces (remember we need u , v , and du ). Note that we don’t
include an arbitrary constant here.

Differentiate u x v e x 
 du 1dx dv e x dx Integrate

Now assemble the integral equation u dv uv  v du from the parts table.
x
x e dx  x e x  e x 1 dx
x
x e dx  x e x  e x  C




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