How do you integrate #(x-5)/(x-2)^2# using partial fractions?

1 Answer
Roy E.
Dec 19, 2016

#ln|x-2|+3/(x-2)+C#

Explanation:

#int(x-5)/(x-2)^2dx#
#=int(x-2-3)/(x-2)^2dx#
#=int(cancel(x-2)/(x-2)^cancel2-3/(x-2)^2)dx#
#=int(1/(x-2)-3/(x-2)^2)dx#
#=ln|x-2|+3/(x-2)+C#.

If the last step is not obvious, substitute #u=x-2#. Alternatively, substitute #u=x-2# right at the start to get #int(1/u-3/u^2)du#, but this isn't then really partial fractions. However, it isn't really a partial fractions question because you have only one unique term in the denominator.

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