Question:
Evaluate: $\int \frac{\mathrm{x}^{2}}{1+\mathrm{x}^{3}}$
Solution:
let $1+x^{3}=t$
Differentiating on both sides we get,
$3 x^{2} d x=d t$
$x^{2} d x=\frac{1}{3} d t$
substituting it in $\int \frac{x^{2}}{1+x^{3}} d x$ we get,
$=\int \frac{1}{3 t} d t$
$=\frac{1}{3} \log t+c$
$=\frac{1}{3} \log \left(1+x^{3}\right)+c$