What is the integral of (\sqrt[3]{x})/(sqrt(x)-1) ?

The integral of (\sqrt[3]{x})/(sqrt(x)-1) is

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integral of (\sqrt[3]{x})/(sqrt(x)-1)

Solution steps

Apply u-substitution

: \int \frac{6 u ^{7}}{u ^{3} - 1} d u

= \int \frac{6 u ^{7}}{u ^{3} - 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{u ^{7}}{u ^{3} - 1} d u

Long division

\frac{u ^{7}}{u ^{3} - 1} : u^{4} + u + \frac{u}{u ^{3} - 1}

= 6 \cdot \int u^{4} + u + \frac{u}{u ^{3} - 1} d u

Apply the Sum Rule:

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

= 6 (\int u^{4} d u + \int u d u + \int \frac{u}{u ^{3} - 1} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int u d u = \frac{u ^{2}}{2}

\int \frac{u}{u ^{3} - 1} d u = \frac{1}{3} ln ∣ u - 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 u ^{2} + 4 u}{3} + 23 arctan (\frac{2 u + 1}{3}))

= 6 (\frac{u ^{5}}{5} + \frac{u ^{2}}{2} + \frac{1}{3} ln ∣ u - 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 u ^{2} + 4 u}{3} + 23 arctan (\frac{2 u + 1}{3})))

Substitute back

Simplify

Add a constant to the solution

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What is the integral of (\sqrt[3]{x})/(sqrt(x)-1) ?
The integral of (\sqrt[3]{x})/(sqrt(x)-1) is