Question

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

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

Answer 1

Answer: (C)

$\log (2+\sqrt{5})$

Let $I=\int \frac{\log (t+\sqrt{1+t^2})}{\sqrt{1+t^2}}dt$

Now assume $y=\log (t+\sqrt{1+t^2})\Rightarrow dy=\frac{1}{t+\sqrt{1+t^2}}.(1+\frac{t}{\sqrt{1+t^2}})dt=\frac{1}{\sqrt{1+t^2}}dt$

$\therefore I=\int ydy=\frac{y^2}{2}+C=\frac{{\left(\log \right(t+\sqrt{1+t^2}\left)\right)}^2}{2}+C$

Thus $g\left(t\right)=\log (t+\sqrt{1+t^2})\Rightarrow g\left(2\right)=\log (2+\sqrt{5})$