Question

Integrate the function: $\frac{e^{2x}-1}{e^{2x}+1}$

Answer 1

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

Dividing numerator and denominator by ex , we obtain

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

Let ex +e-x = t

∴ (ex - e-x)dx = dt

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

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

=log|t|+C

= log|ex + e-x| +C