Indefinite Integration — Maths STD 12 Science — Question

$\begin{aligned} & \text { Let } I =\int \sqrt{x^2+a^2} d x \\ & =\int \sqrt{x^2+a^2} \cdot 1 d x\end{aligned}$$\begin{aligned} & =\sqrt{x^2+a^2} \int 1 d x-\int\left[\frac{d}{d x}\left(\sqrt{x^2+a^2}\right) \cdot \int 1 d x\right] d x \\ & =\sqrt{x^2+a^2} \cdot x-\int \frac{2 x}{2 \sqrt{x^2+a^2}} \cdot x d x \\ & =x \cdot \sqrt{x^2+a^2}-\int \frac{\left(x^2+a^2\right)-a^2}{\sqrt{x^2+a^2}} d x\end{aligned}$$\begin{aligned} & =x \cdot \sqrt{x^2+a^2}-\int\left(\frac{x^2+a^2}{\sqrt{x^2+a^2}}-\frac{a^2}{\sqrt{x^2+a^2}}\right) d x \\ & =x \cdot \sqrt{x^2+a^2}-\int \sqrt{x^2+a^2} d x+a^2 \int \frac{1}{\sqrt{x^2+a^2}} d x \\ & \therefore I =x \cdot \sqrt{x^2+a^2}-I+a^2 \log \left|x+\sqrt{x^2+a^2}\right|+c_1 \\ & \therefore 2 I =x \cdot \sqrt{x^2+a^2}+a^2 \log \left|x+\sqrt{x^2+a^2}\right|+c_1 \\ & \therefore I=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|+\frac{c_1}{2}\end{aligned}$$\therefore \int \sqrt{x^2+a^2} d x=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|+c$, where $c=\frac{c_1}{2}$