Question

Integrate the function: $\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$

Answer 1

Let e2x +e-2x = t

∴ (2e2x -2e-2x)dx = dt

$\Rightarrow$ 2(e2x -e-2x)dx = dt

$\Rightarrow \bigg(\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\bigg)dx=\int\frac{dt}{2t}$

= $\frac{1}{2}\int\frac{1}{t}dt$

= $\frac{1}{2}$log|t|+C

= $\frac{1}{2}$log|e2x +e-2x|+C