Answer:
Option D
Explanation:
We have,
$\int\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}dx=\frac{\sqrt{5}}{3} \tan ^{-1}\frac{x}{\sqrt{5}}-\frac{\sqrt{2}}{3} \tan ^{-1}\frac{x}{\sqrt{2}}+c,$
On differentiating both sides , we get
$\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}$
= $\frac{\sqrt{5}}{3}\left(\frac{1}{1+\frac{x^{2}}{5}}\right)\frac{1}{\sqrt{5}}-\frac{\sqrt{2}}{3}\left(\frac{1}{1+\frac{x^{2}}{2}}\right)\times\frac{1}{\sqrt{2}}$
$\Rightarrow$ $\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}=\frac{1}{3}\left(\frac{5}{5+x^{2}}-\frac{2}{2+x^{2}}\right)$
$\Rightarrow$ $\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}=\frac{x^{2}}{(x^{2}+2)(x^{2}+5)}$
$\Rightarrow$ $\frac{2x^{2}}{(2x^{2}+4)(x^{2}+5)}\Rightarrow\alpha=4$
