Evaluate the following integrals:

Let, $1+\mathrm{x}^{3}=\mathrm{t}$

Differentiate both side with respect to $t$

$3 x^{2} \frac{d x}{d t}=1 \Rightarrow x^{2} d x=\frac{d t}{3}$

$y=\frac{1}{3} \int \frac{(t-1)}{\sqrt{t}} d t$

$y=\frac{1}{3} \int \sqrt{t}-\frac{1}{\sqrt{t}} d t$

$y=\frac{1}{3}\left(\frac{2}{3} t^{\frac{3}{2}}-2 \sqrt{t}\right)+c$

Again, put $t=1+x^{3}$

$y=\frac{1}{3}\left(\frac{2}{3}\left(1+x^{3}\right)^{\frac{3}{2}}-2 \sqrt{1+x^{3}}\right)+c$

$y=\frac{2}{9} \sqrt{1+x^{3}}\left(x^{3}-2\right)+c$