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How to integrate x^2/(x^3+1)?

1 Answer

Ananda Dasgupta

Jun 7, 2018

$\frac{1}{3} ln (x^{3} + 1) + C$ $1/3 ln(x^3+1)+C$

Explanation:

Substitute $x^{3} + 1 = u$ $x^3+1 = u$ .

Then $3 x^{2} d x = d u$ $3x^2dx = du$ and our integral

$\int \frac{x^{2}}{x^{3} + 1} d x$ $int x^2/(x^3+1)dx$

becomes

$\frac{1}{3} \int \frac{d u}{u} = \frac{1}{3} ln u + C$ $1/3 int (du)/u = 1/3 ln u+C$

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