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

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

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Answer 1 · 2017-01-22T16:47:03

$\int \frac{x^{3} - 4}{x + 1} d x$ $int(x^3-4)/(x+1)dx$

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

$= \int \frac{{(u - 1)}^{3} - 4}{u} d u$ $=int((u-1)^3-4)/udu$

Expand ${(u - 1)}^{3}$ $(u-1)^3$ :

$= \int \frac{(u^{3} - 3 u^{2} + 3 u - 1) - 4}{u}$ $=int((u^3-3u^2+3u-1)-4)/u$

$= \int (u^{2} - 3 u + 3 - \frac{5}{u}) d u$ $=int(u^2-3u+3-5/u)du$

Integrate term by term:

$= \frac{1}{3} u^{3} - \frac{3}{2} u^{2} + 3 u - 5 ln | u | + C$ $=1/3u^3-3/2u^2+3u-5lnabsu+C$

$= \frac{1}{3} {(x + 1)}^{3} - \frac{3}{2} {(x + 1)}^{2} + 3 (x + 1) - 5 ln | x + 1 | + C$ $=1/3(x+1)^3-3/2(x+1)^2+3(x+1)-5lnabs(x+1)+C$

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

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