What is the integral from 0 to x^2 of 1/(x^3+1) ?

The integral from 0 to x^2 of 1/(x^3+1) is

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integral from 0 to x^2 of 1/(x^3+1)

\int_{0}^{x^{2}} \frac{1}{x ^{3} + 1} d x

\frac{1}{3} ln x^{2} + 1 + \frac{1}{3} (- \frac{1}{2} ln \frac{1}{3} (2 x^{2} - 1)^{2} + 1 + \frac{1}{2} ln (\frac{4}{3}) + 3 arctan (\frac{1}{3} (2 x^{2} - 1)) + \frac{π}{2 3})

Solution steps

\int_{0}^{x^{2}} \frac{1}{x ^{3} + 1} d x

Take the partial fraction of

\frac{1}{x ^{3} + 1} : \frac{1}{3 ( x + 1 )} + \frac{- x + 2}{3 ( x ^{2} - x + 1 )}

= \int_{0}^{x^{2}} \frac{1}{3 ( x + 1 )} + \frac{- x + 2}{3 ( x ^{2} - x + 1 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{x^{2}} \frac{1}{3 ( x + 1 )} d x + \int_{0}^{x^{2}} \frac{- x + 2}{3 ( x ^{2} - x + 1 )} d x

\int_{0}^{x^{2}} \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln x^{2} + 1

\int_{0}^{x^{2}} \frac{- x + 2}{3 ( x ^{2} - x + 1 )} d x = \frac{1}{3} (- \frac{1}{2} ln \frac{1}{3} (2 x^{2} - 1)^{2} + 1 + \frac{1}{2} ln (\frac{4}{3}) + 3 arctan (\frac{1}{3} (2 x^{2} - 1)) + \frac{π}{2 3})

= \frac{1}{3} ln x^{2} + 1 + \frac{1}{3} (- \frac{1}{2} ln \frac{1}{3} (2 x^{2} - 1)^{2} + 1 + \frac{1}{2} ln (\frac{4}{3}) + 3 arctan (\frac{1}{3} (2 x^{2} - 1)) + \frac{π}{2 3})

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What is the integral from 0 to x^2 of 1/(x^3+1) ?
The integral from 0 to x^2 of 1/(x^3+1) is