Question

Evaluate: $\int \frac{2zdz}{\sqrt[3]{3z^2+1}}$.

Answer 1

Consider the given integral.

$I=\int \frac{2z}{(z^2+1{)}^{1/3}}$

Let $t=z^2+1$

$dt=2zdz$

Therefore,

$I=\int \frac{1}{(t{)}^{1/3}}dt$

$I=\int (t{)}^{-1/3}dt$

$I=\frac{t^{\frac{-1}{3}+1}}{\frac{-1}{3}+1}+C$

$I=\frac{3t^{\frac{2}{3}}}{2}+C$

On putting the value of $t$, we get

$I=\frac{3(z^2+1{)}^{\frac{2}{3}}}{2}+C$

Hence, this is the answer.