Discussion Forum

 If   $\int\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}dx=\frac{\sqrt{5}}{3} \tan ^{-1}\frac{x}{\sqrt{5}}-\frac{\sqrt{2}}{3} \tan ^{-1}\frac{x}{\sqrt{2}}+c,$ then $\alpha$ =

Answer:

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

$\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}$

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

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

$\Rightarrow$     $\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}=\frac{x^{2}}{(x^{2}+2)(x^{2}+5)}$

 $\Rightarrow$      $\frac{2x^{2}}{(2x^{2}+4)(x^{2}+5)}\Rightarrow\alpha=4$