Integral Calculus Question 198
Question: If $ \int{\frac{xe^{x}}{\sqrt{1+e^{x}}}dx=f(x)\sqrt{1+e^{x}}-2\log g(x)+C,} $ then
Options:
A) $ f(x)=x-1 $
B) $ g(x)=\frac{\sqrt{1+e^{x}}-1}{\sqrt{1+e^{x}}+1} $
C) $ g(x)=\frac{\sqrt{1+e^{x}}+1}{\sqrt{1+e^{x}}-1} $
D) $ f(x)=2(2-x) $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ I=\int{\frac{xe^{x}}{\sqrt{1+e^{x}}}dx} $ we have $ \int{\frac{e^{x}}{\sqrt{1+e^{x}}}dx=2\sqrt{1+e^{x}}} $ Integrating I by parts with x as first function $ I=x.2\sqrt{1+e^{x}}-\int{2\sqrt{1+e^{x}}dx} $ $ =2x\sqrt{1+e^{x}}-2\int{t\cdot \frac{2tdt}{t^{2}-1}(Putting1+e^{x}=t^{2})} $ $ =2x\sqrt{1+e^{x}}-4\int{\frac{t^{2}-1+1}{t^{2}-1}dt} $ $ =2x\sqrt{1+e^{x}}-4[ t+\frac{1}{2}\log \frac{t-1}{t+1} ]+c $ $ =2(x-2)\sqrt{1+e^{x}}-2\log ( \frac{\sqrt{1+e^{x}}-1}{\sqrt{1+e^{x}}+1} )+c $
BETA
