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

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

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

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

- \frac{1}{3} ln (2) + \frac{π}{3 3}

Solution steps

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

Take the partial fraction of

\frac{x}{x ^{3} + 1} : - \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}

\frac{d}{d x} (\frac{x - 6}{x ^{2} - 2 x + 12})

f (x) = cos^{2} (x)

x \to 0 lim (- 2 sin (x))

\frac{\partial}{\partial y} (x^{4} y + x z - y z^{2})

\int (\frac{1}{( x ^{2} + 1 ) ^{2}}) d x

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