Integral of $$$\frac{1}{a - x^{2}}$$$ with respect to $$$x$$$

Find $$$\int \frac{1}{a - x^{2}}\, dx$$$.

Let $$$u=x \sqrt{- \frac{1}{a}}$$$.

Then $$$du=\left(x \sqrt{- \frac{1}{a}}\right)^{\prime }dx = \sqrt{- \frac{1}{a}} dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{\sqrt{- \frac{1}{a}}}$$$.

Therefore,

$${\color{red}{\int{\frac{1}{a - x^{2}} d x}}} = {\color{red}{\int{\frac{\sqrt{- a}}{a \left(u^{2} + 1\right)} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{\sqrt{- a}}{a}$$$ and $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$:

$${\color{red}{\int{\frac{\sqrt{- a}}{a \left(u^{2} + 1\right)} d u}}} = {\color{red}{\frac{\sqrt{- a} \int{\frac{1}{u^{2} + 1} d u}}{a}}}$$

The integral of $$$\frac{1}{u^{2} + 1}$$$ is $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$\frac{\sqrt{- a} {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{a} = \frac{\sqrt{- a} {\color{red}{\operatorname{atan}{\left(u \right)}}}}{a}$$

Recall that $$$u=x \sqrt{- \frac{1}{a}}$$$:

$$\frac{\sqrt{- a} \operatorname{atan}{\left({\color{red}{u}} \right)}}{a} = \frac{\sqrt{- a} \operatorname{atan}{\left({\color{red}{x \sqrt{- \frac{1}{a}}}} \right)}}{a}$$

Therefore,

$$\int{\frac{1}{a - x^{2}} d x} = \frac{\sqrt{- a} \operatorname{atan}{\left(x \sqrt{- \frac{1}{a}} \right)}}{a}$$

Simplify:

$$\int{\frac{1}{a - x^{2}} d x} = - \frac{\operatorname{atan}{\left(x \sqrt{- \frac{1}{a}} \right)}}{\sqrt{- a}}$$

Add the constant of integration:

$$\int{\frac{1}{a - x^{2}} d x} = - \frac{\operatorname{atan}{\left(x \sqrt{- \frac{1}{a}} \right)}}{\sqrt{- a}}+C$$

$$$\int \frac{1}{a - x^{2}}\, dx = - \frac{\operatorname{atan}{\left(x \sqrt{- \frac{1}{a}} \right)}}{\sqrt{- a}} + C$$$A