Question

$\int x.\frac{ln\left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}}$ equals :

Answer 1

$\int x.\frac{ln\left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}}$
$\int ln\left(x+\sqrt{1+x^2}\right).\frac{1}{\sqrt{1+x^2}}$
$u=ln\left(x+\sqrt{1+x^2}\right)$
$dv=\frac{1}{x+\sqrt{1+x^2}}\left(1+\frac{1}{2\sqrt{1+x^2}}\times 2x\right)$
$dv=\frac{1}{x+\sqrt{1+x^2}}\left(1+\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right)$
$dv=\frac{1}{\sqrt{1+x^2}}dx$
$\Rightarrow \int udv=\frac{u^2}{2}$
$\Rightarrow \frac{ln\left(x+\sqrt{1+x^2}\right)}{2}+c$