Question

$\underset{-1}{\overset{1}{\int }}\frac{e^x+1}{e^x-1}dx$ equals-

• A. $\ln \left(e^x+1\right)$
• B. $\ln \left(e^x-1\right)$
• C. $1$
• D. 0

Answer 1

The correct option is D 0

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

$thenf(-x)=\left(\frac{e^{-x}+1}{e^{-x}-1}\right)$

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

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

$\therefore f\left(x\right)=Oddfunction$

$\underset{-1}{\overset{1}{\int }}\frac{e^x+1}{e^x-1}dx=0$