Question

$\int \sqrt{a^2+x^2}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \underset{\mathrm{II}}{1·}\underset{\mathrm{I}}{\sqrt{a^2+x^2}}dx \\ =\sqrt{a^2+x^2}\int 1dx-\int \left(\frac{d}{dx}\left(\sqrt{a^2+x^2}\right)\int 1dx\right)dx \\ =\sqrt{a^2+x^2}·x-\int \frac{1\times 2x}{2\sqrt{a^2+x^2}}·xdx \\ =\sqrt{a^2+x^2}·x-\int \left(\frac{x^2+a^2-a^2}{\sqrt{a^2+x^2}}\right)dx \\ =x\sqrt{a^2+x^2}-\int \sqrt{a^2+x^2}dx+a^2\int \frac{1}{\sqrt{a^2+x^2}}dx \\ =x\sqrt{a^2+x^2}-I+a^2\int \frac{1}{\sqrt{a^2+x^2}}dx \\ \therefore 2I=x\sqrt{a^2+x^2}+a^2\ln \left|x+\sqrt{x^2+a^2}\right| \\ \Rightarrow I=\frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\ln \left|x+\sqrt{x^2+a^2}\right|+C \end{gathered}$$