We have, $\int \frac{2x^2}{\left(2x^2+\alpha \right)\left(x^2+5\right)}dx=\frac{\sqrt{5}}{3}{\tan }^{-1}\frac{x}{\sqrt{5}}-\frac{\sqrt{2}}{3}{\tan }^{-1}\frac{x}{\sqrt{2}}+c$

Differentiate both sides w.r.t. $x$, we get, 

$\frac{2x^2}{\left(2x^2+\alpha \right)\left(x^2+5\right)}=\frac{\sqrt{5}}{3}·\frac{1}{1+{\displaystyle \frac{x^2}{5}}}·\frac{1}{\sqrt{5}}-\frac{\sqrt{2}}{3}·\frac{1}{1+{\displaystyle \frac{x^2}{2}}}·\frac{1}{\sqrt{2}}$

Now, the right-hand side 

$=\frac{5}{3}·\frac{1}{x^2+5}-\frac{4}{3}·\frac{1}{2x^2+4}$

$=\frac{1}{3}\left(\frac{10x^2+20-4x^2-20}{\left(x^2+5\right)\left(2x^2+4\right)}\right)$

$=\left(\frac{2x^2}{\left(x^2+5\right)\left(2x^2+4\right)}\right)$.

Hence, we get $\alpha =4$.