How do you evaluate the integral $int (x^3-4)/(x+1)$ ?

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Answer 1 · 2017-01-22T16:47:03

$int(x^3-4)/(x+1)dx$

Let $u=x+1$ , implying that $du=dx$ . This also means that $x=u-1$ . Then:

$=int((u-1)^3-4)/udu$

Expand $(u-1)^3$ :

$=int((u^3-3u^2+3u-1)-4)/u$

$=int(u^2-3u+3-5/u)du$

Integrate term by term:

$=1/3u^3-3/2u^2+3u-5lnabsu+C$

$=1/3(x+1)^3-3/2(x+1)^2+3(x+1)-5lnabs(x+1)+C$

How do you evaluate the integral int (x^3-4)/(x+1)int (x^3-4)/(x+1)?