How do you find the indefinite integral of $int (2x)/(x-1)^2$ ?

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Answer 1 · 2017-02-01T16:19:47

$2 ln |x-1| - 2/(x-1) +c$

Substitute $u=x-1$ . Then $dx=du$ , $x=u+1$ :

$int(2(u+1))/u^2 du$
$=2int 1/u du + 2 int (du)/u^2$
$= 2ln|u| -2/u +c$
$=2ln|x-1|-2/(x-1)+c$

How do you find the indefinite integral of int (2x)/(x-1)^2int (2x)/(x-1)^2?