Answer
$$\int \frac{d x}{\left(x^{2}+a\right)^{2}}=\frac{1}{2 a}\left(\frac{x}{x^{2}+a}+\frac{1}{\sqrt{a}} \tan ^{-1} \frac{x}{\sqrt{a}}\right)+C$$
Work Step by Step
Given $$\int \frac{d x}{\left(x^{2}+a\right)^{2}} $$ Let $$x=\sqrt{a}\tan u\ \ \ \ \ \to \ \ \ \ dx=\sqrt{a}\sec^2 u du$$ Then \begin{align*} \int \frac{d x}{\left(x^{2}+a\right)^{2}}&=\int \frac{\sqrt{a}\sec^2 u du}{\left(a\tan^{2}u+a\right)^{2}}\\ &=\int \frac{\sqrt{a}\sec^2 u du}{ a^2\sec^{4}u}\\ &=\frac{1}{a\sqrt{a}} \int \cos^2 udu\\ &=\frac{1}{2a\sqrt{a}} \int (1+\cos2 u)du\\ &=\frac{1}{2a\sqrt{a}} \left(u+\frac{1}{2}\sin 2u\right)+C\\ &= \frac{1}{2a\sqrt{a}} \left(u+ \sin u\cos u\right)+C\\ &= \frac{1}{2a\sqrt{a}} \left(\frac{\sqrt{a}}{\sqrt{x^{2}+a}} \cdot \frac{x}{\sqrt{x^{2}+a}}+\tan ^{-1} \frac{x}{\sqrt{a}}\right)+C\\ &= \frac{1}{2 a}\left(\frac{x}{x^{2}+a}+\frac{1}{\sqrt{a}} \tan ^{-1} \frac{x}{\sqrt{a}}\right)+C \end{align*}
