Mathematics - Indefinite Integration Question with Solution | TestHub

Put,$\log \left(t+\sqrt{1+t^2}\right)=z$

$\Rightarrow \frac{1}{\left(t+\sqrt{1+t^2}\right)}\left(1+\frac{2t}{2\sqrt{1+t^2}}\right)dt=dz$

$\Rightarrow \frac{\left(t+\sqrt{1+t^2}\right)}{\left( t+\sqrt{1+t^2} \right)}\frac{1}{\sqrt{1+t^2}}dt=dz$

$\Rightarrow \frac{1}{\sqrt{1+t^2}}dt=dz$

$\therefore I=\int zdz=\frac{z^2}{2}+c=\frac{1}{2}{\left[\log \left(t+\sqrt{1+t^2}\right)\right]}^2+c$, where$c$is the constant of integration.

$\therefore g\left(t\right)=\log \left(t+\sqrt{1+t^2}\right)$

$\Rightarrow g\left(2\right)=\log \left(2+ \sqrt{1+4}\right)$

Hence,$\log \left(2+\sqrt{5}\right)$, is the answer.