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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

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