Evaluate the following integrals:

Solution:

Assume $x^{3}+1=t^{2}$

$d\left(x^{3}+1\right)=d\left(t^{2}\right)$

$3 x^{2} \cdot d x=2 t \cdot d t$

$\Rightarrow \mathrm{dx}=\frac{2 \mathrm{t}}{3 \mathrm{x}^{2}} \mathrm{dt}$

$x^{3}+1=t^{2}$

$\Rightarrow \mathrm{dx}=\frac{2 t}{3 \mathrm{x}^{2}} \mathrm{dt}$

Substituting $\mathrm{t}$ and $\mathrm{dt}$

$\Rightarrow \int \frac{x^{5}}{\sqrt{t^{2}}} \cdot \frac{2 t}{3 x^{2}} d t$

$\Rightarrow \int \frac{x^{3}}{t} \cdot \frac{2 t}{3} d t$

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

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

$\Rightarrow \frac{2}{3} \int\left(\mathrm{t}^{2}-1\right) \cdot \mathrm{dt}$

$\Rightarrow \frac{2}{3} \int \mathrm{t}^{2} \mathrm{dt}-\frac{2}{3} \int \mathrm{dt}$

$\Rightarrow \frac{2}{3} \times \frac{\mathrm{t}^{2}}{3}-\frac{2}{3} \mathrm{t}+\mathrm{c}$

$\Rightarrow \frac{2}{9}\left(x^{3}+1\right)^{3 \backslash 2}-\frac{2}{3}\left(x^{3}+1\right)^{1 \backslash 2}+c$