Question

If $\int \frac{\log\left(t+\sqrt{1+t^{2}}\right)}{\sqrt{1+t^{2}}} dt=\frac{1}{2}\left(g\left(t\right)^{2}\right)+C,$ where $C$ is a constant, then $g(2)$ is equal to :

• A. $2\, \log \left(2+\sqrt{5}\right)$
• B. $\log \, \left(2+\sqrt{5}\right)$
• C. $\frac{1}{\sqrt{5}}\, \log \, \left(2+\sqrt{5}\right)$
• D. $\frac{1}{2}\, \log \, \left(2+\sqrt{5}\right)$

Answer 1

Answer: (B)

$I=\int \frac{\log \left(t+\sqrt{1+t^{2}}\right)}{\sqrt{1+t^{2}}} d t$
$\because \frac{d}{d t}\left(\log \left(t+\sqrt{1+t^{2}}\right)\right)=\frac{1}{\sqrt{1+t^{2}}}$
$\Rightarrow I=\frac{1}{2}\left[\log \left(t+\sqrt{ \left.1+t^{2}\right)}\right]^{2}+C\right.$
$\Rightarrow g(t)=\log \left(t+\sqrt{1+t^{2}}\right)$
$\Rightarrow g(2)=\log (2+\sqrt{5})$