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

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

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

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

\frac{1}{3} ln (2) + \frac{π}{3 3}

Solution steps

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

Take the partial fraction of

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

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

= \frac{1}{3} ln (2) + \frac{π}{3 3}

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

\int_{2 y}^{2} 1 d x

\int_{- 2}^{4} \frac{1}{x} d x

\int_{4.2}^{9.9} (12 x^{2} - 6 x + 4) d x

\int_{0}^{4} \frac{1}{2} x d x

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