Definite Integral is not hard!

A simple substitution

Now let $\large{\frac{1}{1+x^3 }=u}$

$\large{x=\left( \frac{1-u}{u} \right )^{\frac{1}{3}}}$

$\large{\frac{x^2 }{(1+x^3 )^2} dx=-\frac{1}{3}du}$

The integral now reduces to

$\large{I=\frac{1}{3} \displaystyle \int^{1}_{0} u^{\frac{1}{3}}(1-u)^{-\frac{1}{3}} du}$

Using

$\large{1) \displaystyle \int^{1}_{0} t^{x-1} (1-t)^{y-1} dt=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}}$

$\large{2) \text{Euler reflection formula}~ \Gamma(z) \Gamma(1-z)=\frac{\pi}{\sin(\pi z)}}$

$\large{3) \Gamma(1+z)=z\Gamma(z)}$

here $x=\frac{4}{3}$ and $y=\frac{2}{3}$

Thus we have

$\large{I=\frac{1}{3} \frac { \Gamma \left( \frac { 4 }{ 3 } \right) \Gamma \left( \frac { 2 }{ 3 } \right) }{ \Gamma \left( \frac { 4+2 }{ 3 } \right) } }$

$\large{I=\frac{1}{3} \times \frac { 1 }{ 3 } \frac { \Gamma \left( \frac { 1 }{ 3 } \right) \Gamma \left( \frac { 2 }{ 3 } \right) }{ \Gamma \left( 2 \right) } }$

$\large{I=\frac{1}{9} \times \frac { \pi }{ \sin { \left( \frac { \pi }{ 3 } \right) } } }$ from $2)$

$\large{I=\boxed{\frac{2 \sqrt{3}}{27} \pi}}$