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

The integral from 0 to 2 of 1/(x^3+1) is (ln(3)+sqrt(3)pi)/6

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

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

\frac{ln ( 3 ) + 3 π}{6}

Decimal Notation

1.09000 \dots

Solution steps

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

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

\int_{0}^{2} \frac{- x + 2}{3 ( x ^{2} - x + 1 )} d x = \frac{- ln ( 3 ) + 3 π}{6}

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

Simplify

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

= \frac{ln ( 3 ) + 3 π}{6}

\int_{0}^{3} 2 x^{2} - 6 x d x

\int_{5}^{9} 2400 x 3 + 2 x d x

\int_{0}^{4} (x^{2} - 2 y^{2} + 1) d x

\int_{2}^{5} 5^{5 x} d x

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

What is the integral from 0 to 2 of 1/(x^3+1) ?
The integral from 0 to 2 of 1/(x^3+1) is (ln(3)+sqrt(3)pi)/6