Question

If $\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}}dx=gof\left(x\right)+$ const.
then

• A. $f\left(x\right)=\log \left(x+\sqrt{x^2+1}\right)$
• B. $f\left(x\right)=\log \left(x+\sqrt{x^2+1}\right)$ and $g\left(x\right)=x^2$
• C. $f\left(x\right)=\log \left(x+\sqrt{x^2+1}\right)$ and $g\left(x\right)=\frac{x^2}{2}$
• D. $f\left(x\right)=\frac{x^2}{2}$ and $g\left(x\right)=\log \left(x+\sqrt{x^2+1}\right)$

Answer 1

The correct option is C $f\left(x\right)=\log \left(x+\sqrt{x^2+1}\right)$ and $g\left(x\right)=\frac{x^2}{2}$

$I=\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}}dx$

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

$\frac{1}{x+\sqrt{1+x^2}}\left(1+\frac{x}{\sqrt{1+x^2}}\right)dx=dt$

$=\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}\left(x+\sqrt{1+x^2}\right)}dx=dt$

$\Rightarrow \frac{1}{\sqrt{1+x^2}}dx=dt$

$I=\int tdt=\frac{1}{2}{t}^2+c$

$=\frac{1}{2}{\left[\log \left(x+\sqrt{1+x^2}\right)\right]}^2+c$

$\left(c\right)option,g=\frac{x^2}{2}$