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

$$\begin{gathered} \int \frac{2{\mathrm{x}}^2}{(2{\mathrm{x}}^2+\mathrm{\alpha }) ({\mathrm{x}}^2+5)}\mathrm{dx} \\ =\frac{\sqrt{5}}{3} {\tan }^{-1} \frac{\mathrm{x}}{\sqrt{5}}-\frac{\sqrt{2}}{3}{\tan }^{-1}\frac{\mathrm{x}}{\sqrt{2}}+\mathrm{c} \end{gathered}$$

differentiate w.r.t 'x'

$$\begin{gathered} \frac{2{\mathrm{x}}^2}{(2{\mathrm{x}}^2+\mathrm{\alpha }) ({\mathrm{x}}^2+5)}=\frac{\sqrt{5}}{3}\frac{1}{1+{\displaystyle \frac{{\mathrm{x}}^2}{5}}}\times \frac{1}{\sqrt{5}} \\ \frac{-\sqrt{2}}{3}\frac{1}{1+{\displaystyle \frac{{\mathrm{x}}^2}{2}}}\times \frac{1}{\sqrt{2}} \\ =\frac{5}{3(5+{\mathrm{x}}^2)}-\frac{2}{3(2+{\mathrm{x}}^2)} \\ =\frac{10+5{\mathrm{x}}^2-10-2{\mathrm{x}}^2}{3(5+{\mathrm{x}}^2) (2+{\mathrm{x}}^2)} \\ =\frac{3{\mathrm{x}}^2}{3(5+{\mathrm{x}}^2) (2+{\mathrm{x}}^2)} =\frac{{\mathrm{x}}^2}{(5+{\mathrm{x}}^2) (2+{\mathrm{x}}^2)} \\ =\frac{2{\mathrm{x}}^2}{(2{\mathrm{x}}^2+4) ({\mathrm{x}}^2+5)} \\ \therefore \mathrm{\alpha }=4 \end{gathered}$$