Question

integrate e^2x-e^-2x/e^2x+e^-2x

Answer 1

$\frac{1}{2} \ln(e^{2x} + e^{-2x}) + C$

Answer 2

The integral $\int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} dx$ is solved by substitution.

Let $u = e^{2x} + e^{-2x}$. Then $du = (2e^{2x} - 2e^{-2x}) dx = 2(e^{2x} - e^{-2x}) dx$.

This implies $(e^{2x} - e^{-2x}) dx = \frac{1}{2} du$.

Substituting these into the integral gives $\int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| + C$.

Replacing $u$ with $e^{2x} + e^{-2x}$ gives the final answer $\frac{1}{2} \ln(e^{2x} + e^{-2x}) + C$.