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

$\log |e^x+e^{-x}|+C$: C is an arbitrary constant

The correct answer is Option (1) → $\log |e^x+e^{-x}+C$: C is an arbitrary constant

Given integral:

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

Let $t = e^{x} \Rightarrow e^{2x} = t^2,\ dx = \frac{dt}{t}$

Substitute in the integral:

$\int \frac{t^2 - 1}{t^2 + 1} \cdot \frac{1}{t} \, dt = \int \frac{t^2 - 1}{t(t^2 + 1)} \, dt$

Use partial fractions:

$\frac{t^2 - 1}{t(t^2 + 1)} = \frac{A}{t} + \frac{Bt + C}{t^2 + 1}$

Multiply both sides by $t(t^2 + 1)$:

$t^2 - 1 = A(t^2 + 1) + (Bt + C)t$

$= A(t^2 + 1) + Bt^2 + Ct$

$= (A + B)t^2 + Ct + A$

Compare coefficients:

• $t^2$: $A + B = 1$
• $t$: $C = 0$
• Constant: $A = -1$

Solve: $A = -1,\ B = 2,\ C = 0$

So the integral becomes:

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

$= -\ln |t| + \ln (t^2 + 1) + C$

$= \ln \left| \frac{t^2 + 1}{t} \right| + C$

Substitute $t = e^x$:

$= \ln \left| \frac{e^{2x} + 1}{e^x} \right| + C$

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