Question

If $\int \frac{x{e}^x}{\sqrt{1+e^x}}dx=f\left(x\right)\sqrt{1+e^x}-2\log g\left(x\right)+C$, then

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

Answer 1

Answer: (B), (C)

$f\left(x\right)=2(x-2)$
$g\left(x\right)=\frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}$

Substituting $1+e^x=t^2\Rightarrow e^{x}dx=2t,x=\log (t^2-1)$
$\int \frac{x{w}^x}{\sqrt{1+e^x}}dx=\int \left(\frac{\log (t^2-1)}{t}\right)2tdt=2\int \log (t^2-1)dt=2[t\log (t^2-1)-2\int \frac{t^2}{t^2-1}dt]+c=2[t\log (t^2-1)-2(1+\frac{1}{2}\log \frac{t-1}{t+1})]+c=2x\sqrt{1+e^x}-4\sqrt{1+e^x}-2\log \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}+c$