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

The integral from 1 to 2 of 1/(1+x^3) is (sqrt(3)ln(3)+pi-2sqrt(3)ln(2))/(6sqrt(3))

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

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

\frac{3 ln ( 3 ) + π - 2 3 ln ( 2 )}{6 3}

Solution steps

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

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

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

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

Simplify

\frac{1}{3} (ln (3) - ln (2)) + \frac{1}{3} (- \frac{1}{2} ln (3) + \frac{π}{2 3}) : \frac{3 ln ( 3 ) + π - 2 3 ln ( 2 )}{6 3}

= \frac{3 ln ( 3 ) + π - 2 3 ln ( 2 )}{6 3}

\int_{2}^{4} x d x

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What is the integral from 1 to 2 of 1/(1+x^3) ?
The integral from 1 to 2 of 1/(1+x^3) is (sqrt(3)ln(3)+pi-2sqrt(3)ln(2))/(6sqrt(3))