Question

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

Answer 1

Let $I=\int \log (x+\sqrt{x^2+a^2})dx$

$=\int 1.\log (x+\sqrt{x^2+a^2})dx$

applying $ILATE$ rule

Let $u=\log (x+\sqrt{x^2+a^2})$ and $v=1$

$\int uvdx=u\int vdx-\int \left[\frac{du}{dx}\int vdx\right]dx$

$=\log (x+\sqrt{x^2+a^2})\int 1.dx-\int \left[\int \frac{d}{dx}\left(\log (x+\sqrt{x^2+a^2})\right)1.dx\right]dx$

$=x\log (x+\sqrt{x^2+a^2})-\frac{1}{2}\int \frac{2x}{\sqrt{x^2+a^2}}dx$

Let $t=x^2+a^2\Rightarrow dt=2xdx$

$=x\log (x+\sqrt{x^2+a^2})-\frac{1}{2}\int \frac{dt}{\sqrt{t}}$

$=x\log (x+\sqrt{x^2+a^2})-\frac{1}{2}\int t^{-\frac{1}{2}}dt$

$=x\log (x+\sqrt{x^2+a^2})-\frac{1}{2}\frac{t^{-\frac{1}{2}+1}}{\frac{-1}{2}+1}+c$

$=x\log (x+\sqrt{x^2+a^2})-\frac{1}{2}\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c$

$=x\log (x+\sqrt{x^2+a^2})-\sqrt{t}+c$

$=x\log (x+\sqrt{x^2+a^2})-\sqrt{x^2+a^2}+c$ where $t=x^2+a^2$