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

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

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

Solution steps

Apply u-substitution

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

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

Long division

\frac{u ^{7}}{1 + u ^{3}} : 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 4 u^{2} - 4 u + 4 + 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 4 u^{2} - 4 u + 4 + 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})/(1+sqrt(x)) ?
The integral of (\sqrt[3]{x})/(1+sqrt(x)) is