Question

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

Answer 1

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

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

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

Consider:

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

Assuming $e^{2x}+1=t$

$\Rightarrow 2e^{2x}dx=dt$

$\Rightarrow 2dx=\frac{dt}{t-1}$

Now,

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

$E=\ln (t-1)-\ln \left(t\right)+k$, where $k$ is the constant of integration

$\Rightarrow E=\ln \left(\frac{t-1}{t}\right)+k$

$\Rightarrow E=\ln \left(\frac{e^{2x}}{e^{2x}+1}\right)+k$

Therefore,

$I=\int \frac{e^{2x}-1}{e^{2x}+1}dx=x-\ln \left(\frac{e^{2x}}{e^{2x}+1}\right)+k$