Solving the indefinite integral $\int\frac{dx}{(e^x + e^{-x})^2}$ by conjugates

Question

Evaluate

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

For one method - the first step to take would be to multiply by conjugate

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

My question is: Why would the conjugate of $(e^x + e^{-x})^2$ be? $\dfrac{(e^x)^2}{(e^x)^2}$?

Taken from my professor.

Answer 1

$e^{2x}$ is Not conjugate of $(e^x + e^{-x})^2$

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

Now making an obvious substitution $$e^{2x}+1=t\iff 2e^{2x}\,\mathrm dx=\,\mathrm dt$$

$$I=\int\frac{\,\mathrm dt}{2t^2}=-\frac{1}{2t}+c=-\frac{1}{2e^{2x}+2}+c$$

$$\large\int\frac{\,\mathrm dx}{(e^x + e^{-x})^2}=-\frac{1}{2e^{2x}+2}+c$$