Question

Prove that $\int \frac{e^{2x}-1}{e^{2x}+1}dx$

• A. $\log (e^x-e^{-x})+c$
• B. $\log (e^x+e^{-x})+c$
• C. $\log (2e^x+e^{-x})+c$
• D. None of these

Answer 1

Answer: (B)

$\log (e^x+e^{-x})+c$

Integrating the function

Let $I=\int \frac{e^{2x}-1}{e^{2x}+1}.dx$

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

Putting $e^x+e^{-x}=t$ & $dx=\frac{dt}{e^x-e^{-x}}$

$=\int \frac{e^x-e^{-x}}{t}.\frac{dt}{e^x-e^{-x}}$

$=\int \frac{1}{t}.dt$

$=\log |t|+C$

$=\log |e^x+e^{-x}|+C$ (Using $t=e^x+e^{-x}$)

$I=\log (e^x+e^{-x})+C$ (As $e^x+e^{-x}>0$)