What is the integral of x/((x^2+1)^{1/3)} ?

The integral of x/((x^2+1)^{1/3)} is 3/4 (x^2+1)^{2/3}+C

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integral of x/((x^2+1)^{1/3)}

integral

\frac{x}{( x ^{2} + 1 ) ^{\frac{1}{3}}}

\frac{3}{4} (x^{2} + 1)^{\frac{2}{3}} + C

Solution steps

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{1}{3}}} d x

Apply u-substitution

= \int \frac{1}{2 u ^{\frac{1}{3}}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{\frac{1}{3}}} d u

Apply the Power Rule

= \frac{1}{2} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = x^{2} + 1

= \frac{1}{2} \cdot \frac{3}{2} (x^{2} + 1)^{\frac{2}{3}}

Simplify

\frac{1}{2} \cdot \frac{3}{2} (x^{2} + 1)^{\frac{2}{3}} : \frac{3}{4} (x^{2} + 1)^{\frac{2}{3}}

= \frac{3}{4} (x^{2} + 1)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{4} (x^{2} + 1)^{\frac{2}{3}} + C

What is the integral of x/((x^2+1)^{1/3)} ?
The integral of x/((x^2+1)^{1/3)} is 3/4 (x^2+1)^{2/3}+C