Question

If the primitive of $\frac{e^x(1+e^x)}{\sqrt{1-e^{2x}}}$ is $fog\left(x\right)-\sqrt{h\left(x\right)}+C$ then

• A. $f\left(x\right)=si{n}^{-1}x$
• B. $g\left(x\right)=e^{2x}$
• C. $g\left(x\right)=e^x$
• D. $h\left(x\right)=1-e^{2x}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $g\left(x\right)=e^x$
C $h\left(x\right)=1-e^{2x}$
D $f\left(x\right)=si{n}^{-1}x$

The integrand can be written as $\frac{e^x}{\sqrt{1-e^{2x}}}+\frac{e^{2x}}{\sqrt{1-e^{2x}}}$
So the primitive is equal to
$\int \frac{dt}{\sqrt{1-t^2}}+\frac{1}{2}\int \frac{du}{\sqrt{1-u}}$
By substituting $e^x=t$ and $e^{2x}=u$
$={\sin }^{-1}t-\sqrt{1-u}+c={\sin }^{-1}{e}^x-\sqrt{1-e^{2x}}+c$
Hence $f\left(x\right)={\sin }^{-1}x,g\left(x\right)=e^x$ and $h\left(x\right)=1-e^{2x}$