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

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

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

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

- \frac{1}{3} ln (2) + \frac{π}{3 3}

Decimal

0.37355 \dots

Solution steps

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

Take the partial fraction of

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

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

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

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

= - \frac{1}{3} ln (2) + \frac{π}{3 3}

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