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

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

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

Solution steps

Apply u-substitution

: \int ln (1 + u) \cdot 3 u^{2} d u

= \int ln (1 + u) \cdot 3 u^{2} d u

Take the constant out:

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

= 3 \cdot \int ln (1 + u) u^{2} d u

Apply u-substitution

: \int (v - 1)^{2} ln (v) d v

= 3 \cdot \int (v - 1)^{2} ln (v) d v

Apply Integration By Parts

: ln (v) (\frac{v ^{3}}{3} - v^{2} + v) - \int \frac{1}{3} (v^{2} - 3 v + 3) d v

= 3 (ln (v) (\frac{v ^{3}}{3} - v^{2} + v) - \int \frac{1}{3} (v^{2} - 3 v + 3) d v)

\int \frac{1}{3} (v^{2} - 3 v + 3) d v = \frac{v ^{3}}{9} - \frac{v ^{2}}{2} + v

= 3 (ln (v) (\frac{v ^{3}}{3} - v^{2} + v) - (\frac{v ^{3}}{9} - \frac{v ^{2}}{2} + v))

Substitute back

Simplify

Add a constant to the solution

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