Question

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

• A. $I=\log |e^x+e^{-x}|+C$
• B. $I=2\log |e^x+e^{-x}|+C$
• C. $I=\log |e^x-e^{-x}|+C$
• D. None of these

Answer 1

The correct option is A $I=\log |e^x+e^{-x}|+C$

Let us consider,

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

multiply by $e^{--x}$

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

Let ,

$e^x+e^{-x}=t$

$e^x+(-1)e^{-x}=\frac{dt}{dx}$

From (1)

$=\int \frac{dt}{t}=\log t+C$

$=\log |e^x+e^{-x}|+C$