Question

Integrate $\int \frac{e^x-1}{e^x+1}dx$

• A. $2\log (e^{-x}+1)+x+C$
• B. $2\log (e^x+1)+x+C$
• C. $2\log (e^{-x}+1)-x+C$
• D. $Noneofthese$

Answer 1

The correct option is C $2\log (e^{-x}+1)-x+C$

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

putting $e^x+1=t$

$\Rightarrow e^{x}dx=dt$

$\Rightarrow dx=\frac{dt}{e^x}$

$\Rightarrow dx=\frac{dt}{t-1}$

$={\int }^{}\frac{t-2}{t\left(t-1\right)}dt$

$={\int }^{}\left(\frac{2}{t}-\frac{1}{t-1}\right)dt$

$=2\log t-\log \left(t-1\right)+c$

$=2\log \left(e^x+1\right)-\log \left(e^x\right)+c$

$=2\log \left(e^x+1\right)-x+c$