Question

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

Answer 1

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

$=\int \frac{e^{4x}-1}{e^{4x}+1}dx$

$=\int \frac{e^{4x}+1}{e^{4x}+1}-\frac{2}{e^{4x}+1}dx$

$=\int dx-\int \frac{2}{e^{4x}+1}dx$

Let $e^{4x}+1=t$

$\Rightarrow 4.e^{4}xdx=dt$

$\Rightarrow 4(t-1)dx=dt$

$=x-\int \frac{2}{t}\frac{dt}{4(t-1)}$

$=x-\frac{1}{2}\int \frac{dt}{t(t-1)}$

$=x-\frac{1}{2}\int -(\frac{1}{t}-\frac{1}{t-1})dt$

$=x+\frac{1}{2}\log t-\log (t-1)+c$

$=x+\frac{1}{2}(e^{4x}+1)\log e^{4x}+c.$

Hence, the answer is $x+\frac{1}{2}(e^{4x}+1)\log e^{4x}+c.$