Question

lf 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. $f\left(x\right)={\sin }^{-1}x$
• 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)

$g\left(x\right)=e^x$
$f\left(x\right)={\sin }^{-1}x$
$h\left(x\right)=1-e^{2x}$

$\int \frac{e^x(1+e^x)}{\sqrt{1-e^{2x}}}dx$

$=\int \frac{e^x}{\sqrt{1-e^{2x}}}dx+\int \frac{e^{2x}}{\sqrt{1+e^{2x}}}dx$

$=si{n}^{-1}\left(e^x\right)+\frac{1}{2}\int \frac{d{e}^{2x}}{\sqrt{1-e^{2x}}}$

$=si{n}^{-1}\left(e^x\right)-\sqrt{1-e^{2x}}+c$

So, $g\left(x\right)=e^x$

$f\left(x\right)=si{n}^{-1}x$ and

$h\left(x\right)=1-e^{2x}$