How do I find the integral #intdx/(x^2(x-1)^2)# ?

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Answer 1 · 2014-09-28T11:59:18

By Partial Fraction Decomposition,

#1/{x^2(x-1)^2}=2/x+1/x^2+2/{x-1}+1/{(x-1)^2}#.

#int{dx}/{x^2(x-1)^2}#

by the partial fractions above,

#=int2/x+1/x^2+2/{x-1}+1/{(x-1)^2}dx#

by Log Rule and Power Rule,

#=2ln|x|-1/x+2ln|x-1|-1/{x-1}+C#