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Integration by Parts
Integration by parts is the reverse of the product rule for differentiation.
Suppose we want to integrate the product of two functions, f (x) × g(x).
We need to be able to integrate one of the functions, say f (x): let F (x) be an
antiderivative of f (x), i.e.
Z
f (x) dx = F (x) + C .
Then the Integration by Parts formula is
Z
Z
f (x)g(x) dx = F (x)g(x) − F (x)g 0 (x) dx .
Examples using this formula (These examples have a trigonometric function as f (x) and a power of x or a polynomial as g(x) .)
To remember the formula, note that both terms on the right-hand side involve the integral of f (x), but g(x) is differentiated only in the second term.
Notes
1. Don’t imagine that Integration by Parts is useful for any integral involving
a product. See Example 4a on the link below.
Example where Integration by Parts doesn’t work
2. The formula generates an integral of a new product (on the right-hand
side). It may require one or more further integrations by parts (or the application of other techniques) before we can finally evaluate the integral. See
Example 2 on the link below.
Example requiring 2 integrations by parts
3. If you only know the antiderivative of one of the functions in the product,
then that function obviously has to be “f (x)” in the formula.
An example of a type of product where the antiderivative
of only one of the functions is known
If you know antiderivatives of both the functions in the product, then:
• If there is a positive integer power of x multiplied by another function,
let the power of x be “g(x)” in the formula: since it gets differentiated
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in the integral on the right-hand side of the formula, the n’th power of x
will become a constant after n integrations by parts.
A type of product of a positive integer power with another function which can be antidifferentiated
• In other cases, it usually doesn’t matter which of the functions in the
product you choose as “f (x)” and which as “g(x)”.
A type of product where it doesn’t matter which of the
functions is “f (x)” and which is “g(x)”
4. Sometimes Integration by Parts works for integrals that don’t look like
products at all: suppose we want to integrate g(x), but we only know how
to differentiate g(x). Then let f (x) = 1, so g(x) = f (x).g(x), and use the
Integration by Parts formula.
How to integrate g(x) = ln x , a function which we only
know how to differentiate
5. Most textbooks write the Integration by Parts formula in the form
Z
Z
dv
du
u
dx = uv −
v
dx .
dx
dx
If you are already familiar with using it in this form, there is no need to learn
the form given earlier.
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