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

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Long division got me $\int x^{2} + x + 2 + \frac{2}{x - 1} d x$ $\intx^2+x+2+2/(x-1)dx$

Solving:
$\int \frac{2}{x - 1} d x$ $\int2/(x-1)dx$ by u-sub ( $u = x - 1$ $u=x-1$ and $d u = d x$ $du=dx$ ) becomes $2 \int \frac{1}{u} d u = 2 ln u + C = 2 ln | x - 1 | + C$ $2\int1/udu=2\lnu+C=2\ln|x-1|+C$
$\therefore\intx^2+x+2+2/(x-1)dx=x^3/3+x^2/2+2x+2\ln|x-1|+C$

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Answer 1 · 2018-04-02T18:37:13

$I=x^3/3+x^2/2+2x+2ln|x-1|+c$
i.e.Your answer is O.K.

Here,

$I=int(x^3+x)/(x-1)dx=int(x^3-1+(x-1)+2)/(x-1)dx$

$=int((x-1)(x^2+x+1)+(x-1)+2)/(x-1)dx$

$=int((x^2+x+1)+1)dx+int2/(x-1)dx$

$=x^3/3+x^2/2+x+x+2ln|x-1|+c$

$=x^3/3+x^2/2+2x+2ln|x-1|+c$