Evaluate this integral with substitution

Question

Use the substitution $u=e^x$ and then partial fractions to solve $$\int_1^\infty\frac{e^{-x}}{(e^x+1)(e^x-1)}\ dx$$

Answer 1

Since you are just new on this site, I shall try to help you.

$$I=\int\frac{e^{-x}}{(e^x+1)(e^x-1)}\, dx=\int\frac{dx}{e^x(e^x+1)(e^x-1)}\, dx$$ Change variable $$e^x=t\implies x=\log(t)\implies dx=\frac{dt}t$$ So, $$I=\int \frac{dt}{t^2(t+1)(t-1)}$$ Now, partial fraction decomposition $$\frac{1}{t^2(t+1)(t-1)}=-\frac{1}{t^2}-\frac{1}{2 (t+1)}+\frac{1}{2 (t-1)}$$ Integration of each term $$I=\frac 1t-\frac12\log(t+1)+\frac12\log(t-1)=\frac 1t+\frac12\log\frac{t-1}{t+1}$$ If you want, go back to $x$ $$I=e^{-x}+\frac12\log\left(\frac{e^x-1}{e^x+1}\right)$$ You are certainly able to recognize the last term to be a well know hyperbolic function.

Now, your turn !