Question

The $f\left(x\right)=\sqrt{x},g\left(x\right)=e^x-1\forall x\in (0,\infty )$ and $\int fog\left(x\right)dx=Afog\left(x\right)+B{\tan }^{-1}\left(fog\right(x\left)\right)+C$, then $A+B$ is

Answer 1

$fog\left(x\right)=\sqrt{e^x-1}$
$\int fog\left(x\right)dx=\int \sqrt{e^x-1}dx$
Assuming
$\sqrt{e^x-1}=t\Rightarrow t^2+1=e^x\Rightarrow e^{x}dx=2tdt\Rightarrow dx=\frac{2t}{t^2+1}dt$
$\Rightarrow \int fog\left(x\right)dx=\int \frac{2t^2}{t^2+1}dt\Rightarrow \int fog\left(x\right)dx=2\int [\frac{t^2+1}{t^2+1}-\frac{1}{t^2+1}]dt$
$\Rightarrow \int fog\left(x\right)dx=2\sqrt{e^x-1}-2{\tan }^{-1}\left(\sqrt{e^x-1}\right)+c$

Hence $A+B=0$