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How to integrate quotients where the
numerator is the derivative of the
denominator
To evaluate
Z
g 0 (x)
dx
g(x)
for any function g(x), use the substitution
u = g(x) .
Then
du = g 0 (x) dx ,
so
Z
g 0 (x)
dx =
g(x)
=
=
Z
1
du
u
ln |u| + C
using a standard integral
ln |g(x)| + C
The result
Z
g 0 (x)
dx = ln |g(x)| + C
g(x)
is worth remembering.
More commonly, you will find integrands where the numerator is a numerical
constant multiplied by the derivative of the denominator. To evaluate
Z
h(x)
dx ,
g(x)
if you find that
h(x) = kg 0 (x)
for some number k:
Z
(1)
h(x)
dx = k ln |g(x)| + C .
g(x)
The procedure is to differentiate the denominator and see if (1) is satisfied; if
so, find the number k that satisfies it. See worked example no. 3.
Worked examples of integrating quotients of this type
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Go to List of types of integrand
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2