Question

$\int \log \left(x+\sqrt{x^2+a^2}\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \underset{}{\underset{\mathrm{II}}{1}·\log }\underset{\mathrm{I}}{\left(x+\sqrt{x^2+a^2}\right)}dx \\ =\log \left(x+\sqrt{x^2+a^2}\right)\int 1dx-\int \left[\frac{d}{dx}\left\{\log \left(x+\sqrt{x^2+a^2}\right)\right\}\int 1dx\right] \\ =\log \left(x+\sqrt{x^2+a^2}\right)·x-\int \left(\frac{1}{x+\sqrt{x^2+a^2}}\right)\times \left(1+\frac{1\times 2x}{2\sqrt{x^2+a^2}}\right)·x·dx \\ =\log \left(x+\sqrt{x^2+a^2}\right)·x-\int \frac{x}{\sqrt{x^2+a^2}}dx \\ \mathrm{Putting}{x}^2+a^2=t\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{integral} \\ \Rightarrow 2xdx=dt \\ \Rightarrow xdx=\frac{dt}{2} \\ \therefore I=x·\log \left(x+\sqrt{x^2+a^2}\right)-\frac{1}{2}\int \frac{1}{\sqrt{t}}dt \\ =x·\log \left(x+\sqrt{x^2+a^2}\right)-\frac{1}{2}\int t^{-\frac{1}{2}}dt \\ =x·\log \left(x+\sqrt{x^2+a^2}\right)-\frac{1}{2}\left[\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\right]+C \\ =x·\log \left(x+\sqrt{x^2+a^2}\right)-\sqrt{t}+C \\ =x·\log \left(x+\sqrt{x^2+a^2}\right)-\sqrt{x^2-a^2}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \mathit{}t=x^{\mathit{2}}+a^{\mathit{2}}\right] \end{gathered}$$