Question

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

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

Answer 1

Answer: (C)

$f\left(x\right)=\log (x+\sqrt{x^2+1})$ and $g\left(x\right)=x^2/2$

$\int \frac{log(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$
$log(x+\sqrt{1+x^2})=t\frac{\frac{\sqrt{1+x^2+1}}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}-dx=dt$
$\frac{1}{\sqrt{1+x^2}}dx=dt$
So, $\int t.dt$
$=t^2/2+c$
$=\frac{\left(log\right(x+\sqrt{1+x^2}){)}^2}{2}+c$
$f\left(x\right)=log(x+\sqrt{1+x^2})$
$g\left(x\right)=x^2/2$
Hence, option 'C' is correct.