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

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Answer 1 · 2016-07-29T15:59:53

$int (x-5)/(x-2)^2 .d x=l n(|x-2|)-3/(x-2)+C$

$int (x-5)/(x-2)^2 .d x=?$

$"let us write (x-5) as (x-2-3)"$

$int (x-2-3)/(x-2)^2 * d x$

$"split (x-2-3 )$

$int cancel((x-2))/cancel((x-2)^2)* d x-int 3/(x-2)^2 *d x$

$int (d x)/(x-2)-3 int (d x)/(x-2)^2$

$"1- solve " int (d x)/(x-2)$

$"substitute u=x-2" ; "d u=d x$

$int (d x)/(x-2)=(d u)/u=l n u$

$"undo substitution "$

$int (d x)/(x-2)=l n (x-2)$

$"2- now solve "3 int (d x)/(x-2)^2$

$u=x-2" ; " d x=d u" ; "3 int (d u)/u^2=3int u^-2 d u$

$3int (d x)/(x-2)=1/(-2+1 )u^(-2+1)=-3*u^(-1)=-3/u$

$"undo substitution"$

$3int(d x)/(x-2)^2=-3/(x-2)$

$"The problem is solved:"$

$int (x-5)/(x-2)^2 .d x=l n(|x-2|)-3/(x-2)+C$

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