How do you find the indefinite integral of $int (x^2-6x-20)/(x+5)$ ?

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Answer 1 · 2016-11-30T00:40:10

$int frac (x^2-6x-20) (x+5) dx = (x+5)^2/2 -16(x+5) +35 ln (x+5)$

Substitute $x=t-5$

$int frac (x^2-6x-20) (x+5) dx = int (dt)/t((t-5)^2 -6(t-5) -20) =$

$= int (dt)/t(t^2-10t +25 -6t+30 -20) =$

$= int (dt)/t(t^2-16t +35) = int t dt -16 int dt +35 int (dt)/t =$

$= (t^2)/2 -16t +35 ln(t) = (x+5)^2/2 -16(x+5) +35 ln (x+5)$

How do you find the indefinite integral of int (x^2-6x-20)/(x+5)int (x^2-6x-20)/(x+5)?