What is the integral of arctan(x^{1/3}) ?

The integral of arctan(x^{1/3}) is xarctan(x^{1/3})-1/2 x^{2/3}-1/2+1/2 ln|x^{2/3}+1|+C

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integral of arctan(x^{1/3})

\int arctan (x^{\frac{1}{3}}) d x

x arctan (x^{\frac{1}{3}}) - \frac{1}{2} x^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{2} ln x^{\frac{2}{3}} + 1 + C

Solution steps

\int arctan (x^{\frac{1}{3}}) d x

Apply u-substitution

= \int arctan (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 arctan (u) u^{2} d u

Apply Integration By Parts

= 3 (\frac{1}{3} u^{3} arctan (u) - \int \frac{u ^{3}}{3 ( u ^{2} + 1 )} d u)

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

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

Substitute back

u = x^{\frac{1}{3}}

= 3 (\frac{1}{3} (x^{\frac{1}{3}})^{3} arctan (x^{\frac{1}{3}}) - \frac{1}{6} ((x^{\frac{1}{3}})^{2} + 1 - ln (x^{\frac{1}{3}})^{2} + 1))

Simplify

3 (\frac{1}{3} (x^{\frac{1}{3}})^{3} arctan (x^{\frac{1}{3}}) - \frac{1}{6} ((x^{\frac{1}{3}})^{2} + 1 - ln (x^{\frac{1}{3}})^{2} + 1)) : x arctan (x^{\frac{1}{3}}) - \frac{1}{2} x^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{2} ln x^{\frac{2}{3}} + 1

= x arctan (x^{\frac{1}{3}}) - \frac{1}{2} x^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{2} ln x^{\frac{2}{3}} + 1

Add a constant to the solution

= x arctan (x^{\frac{1}{3}}) - \frac{1}{2} x^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{2} ln x^{\frac{2}{3}} + 1 + C

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What is the integral of arctan(x^{1/3}) ?
The integral of arctan(x^{1/3}) is xarctan(x^{1/3})-1/2 x^{2/3}-1/2+1/2 ln|x^{2/3}+1|+C