If int frac{2 x^2}{left(2 x^2+alpharight)left | vedclass

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(D) Given, $\int \frac{2 x^2}{\left(2 x^2+\alphaight)\left(x^2+5ight)} d x=\frac{\sqrt{5}}{3} 	an ^{-1} \frac{x}{\sqrt{5}}-\frac{\sqrt{2}}{3} 	a

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$4$

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(D) Given, $\int \frac{2 x^2}{\left(2 x^2+\alphaight)\left(x^2+5ight)} d x=\frac{\sqrt{5}}{3} 	an ^{-1} \frac{x}{\sqrt{5}}-\frac{\sqrt{2}}{3} 	an ^{-1} \frac{x}{\sqrt{2}}+c$. Differentiating both sides with respect to $x$, we get: $\frac{2 x^2}{\left(2 x^2+\alphaight)\left(x^2+5ight)} = \frac{\sqrt{5}}{3} \cdot \frac{1}{1+\frac{x^2}{5}} \cdot \frac{1}{\sqrt{5}} - \frac{\sqrt{2}}{3} \cdot \frac{1}{1+\frac{x^2}{2}} \cdot \frac{1}{\sqrt{2}}$. Simplifying the right-hand side: $= \frac{1}{3} \cdot \frac{5}{x^2+5} - \frac{1}{3} \cdot \frac{2}{x^2+2} = \frac{1}{3} \left( \frac{5(x^2+2) - 2(x^2+5)}{(x^2+5)(x^2+2)} ight)$. $= \frac{1}{3} \left( \frac{5x^2 + 10 - 2x^2 - 10}{(x^2+5)(x^2+2)} ight) = \frac{3x^2}{3(x^2+5)(x^2+2)} = \frac{x^2}{(x^2+5)(x^2+2)}$. Comparing this with the left-hand side $\frac{2x^2}{(2x^2+\alpha)(x^2+5)}$, we have: $\frac{2x^2}{(2x^2+\alpha)(x^2+5)} = \frac{x^2}{(x^2+5)(x^2+2)}$. This implies $2(x^2+2) = 2x^2 + \alpha$, so $2x^2 + 4 = 2x^2 + \alpha$. Thus, $\alpha = 4$.

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

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