Question

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

Answer 1

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