What is the integral from 1 to 2 of 1/(x^3+8) ?

The integral from 1 to 2 of 1/(x^3+8) is 1/12 (2ln(2)-ln(3))+1/12 (-1/2 ln(4/3)+pi/(2sqrt(3)))

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integral from 1 to 2 of 1/(x^3+8)

\int_{1}^{2} \frac{1}{x ^{3} + 8} d x

\frac{1}{12} (2 ln (2) - ln (3)) + \frac{1}{12} (- \frac{1}{2} ln (\frac{4}{3}) + \frac{π}{2 3})

Decimal Notation

0.08756 \dots

Solution steps

\int_{1}^{2} \frac{1}{x ^{3} + 8} d x

Take the partial fraction of

\frac{1}{x ^{3} + 8} : \frac{1}{12 ( x + 2 )} + \frac{- x + 4}{12 ( x ^{2} - 2 x + 4 )}

= \int_{1}^{2} \frac{1}{12 ( x + 2 )} + \frac{- x + 4}{12 ( x ^{2} - 2 x + 4 )} 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}{12 ( x + 2 )} d x + \int_{1}^{2} \frac{- x + 4}{12 ( x ^{2} - 2 x + 4 )} d x

\int_{1}^{2} \frac{1}{12 ( x + 2 )} d x = \frac{1}{12} (2 ln (2) - ln (3))

\int_{1}^{2} \frac{- x + 4}{12 ( x ^{2} - 2 x + 4 )} d x = \frac{1}{12} (- \frac{1}{2} ln (\frac{4}{3}) + \frac{π}{2 3})

= \frac{1}{12} (2 ln (2) - ln (3)) + \frac{1}{12} (- \frac{1}{2} ln (\frac{4}{3}) + \frac{π}{2 3})

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What is the integral from 1 to 2 of 1/(x^3+8) ?
The integral from 1 to 2 of 1/(x^3+8) is 1/12 (2ln(2)-ln(3))+1/12 (-1/2 ln(4/3)+pi/(2sqrt(3)))