Answer
$$\int \frac{d x}{x^{2}+a}=\frac{1}{\sqrt{a}} \tan ^{-1} \frac{x}{\sqrt{a}}+C$$
Work Step by Step
Given $$\int \frac{d x}{x^{2}+a} $$ Let $$x=\sqrt{a}\tan u\ \ \ \ \ \to \ \ \ \ dx=\sqrt{a}\sec^2 u du$$ Then \begin{align*} \int \frac{d x}{x^{2}+a} &=\int \frac{\sqrt{a}\sec^2 u du}{a\tan^{2}u+a} \\ &=\int \frac{\sqrt{a}\sec^2 u du}{a\sec^{2}u} \\ &=\frac{1}{\sqrt{a}}\int du\\ &=\frac{1}{\sqrt{a}} u\\ &=\frac{1}{\sqrt{a}} \tan ^{-1} \frac{x}{\sqrt{a}}+C \end{align*}
