Answer
$$\displaystyle{\int_0^a\frac{1}{\left(a^2+x^2\right)^{\frac{3}{2}}}dx=\frac{1}{a^2\sqrt{2}}+C}\\ $$
Work Step by Step
$\displaystyle{I=\int_0^a\frac{1}{\left(a^2+x^2\right)^{\frac{3}{2}}}dx}\\$
$\displaystyle \left[\begin{array}{ll} x=a\tan\theta & x^2=a^2\tan^2\theta \\ & \\ \frac{dx}{d\theta}=a\sec^2\theta & dx=a\sec^2\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int_0^{\frac{\pi}{4}}\frac{1}{\left(a^2+a^2\tan^2\theta\right)^{\frac{3}{2}}}a\sec^2\theta\ d\theta}\\ \displaystyle{I=\int_0^{\frac{\pi}{4}}\frac{1}{\left(a^2\left(1+\tan^2\theta\right)\right)^{\frac{3}{2}}}a\sec^2\theta\ d\theta}\\ \displaystyle{I=\int _0^\frac{\pi }{4}\frac{a\sec^2\theta}{a^3\sec^3\theta}\ d\theta}\\ \displaystyle{I=\int _0^\frac{\pi }{4}\frac{1}{a^2\sec\theta}\ d\theta}\\ \displaystyle{I=\frac{1}{a^2}\int _0^\frac{\pi }{4}\cos\theta\ d\theta}\\ \displaystyle{I=\frac{1}{a^2}\left[\sin\theta\right]_0^\frac{\pi }{4}+C}\\ \displaystyle{I=\frac{1}{a^2}\left(\frac{1}{\sqrt{2}}-0\right)+C}\\ \displaystyle{I=\frac{1}{a^2\sqrt{2}}+C}\\ $
