How do you integrate $x^3/(x-2)$ ?

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Answer 1 · 2015-07-07T02:18:19

Do the division, then integrate to get $x^3/3 + x^2 +4x +8ln abs(x-2) + C$

Perform the division:

$x^3/(x-2) = x^2 + 2x +4 +8/(x-2)$

So
$int x^3/(x-2) dx = int (x^2 + 2x +4 +8/(x-2)) dx$

$= x^3/3 + x^2 +4x +8ln abs(x-2) + C$

How do you integrate x^3/(x-2)x^3/(x-2)?