Integrate $\int_{-\infty}^{-1}\frac{e^x-e^{-x}}{1-e^{-2x}}dx$

Question

I need to solve

$$\int_{-\infty}^{-1}\frac{e^x-e^{-x}}{1-e^{-2x}}dx$$

and recognized that it's $u$-substitution based. I took $u$ as $e^x$ and $du=e^xdx$.

$$\int_{-\infty}^{-1}\frac{e^x-e^{-x}}{1-e^{-2x}}dx=\int_{-\infty}^{-1}\frac{e^x}{1-e^{-2x}}dx+\int_{-\infty}^{-1}\frac{e^{-x}}{1-e^{-2x}}dx$$

$$\int_{-\infty}^{-1}\frac{u}{1-u^2}dx+\int_{-\infty}^{-1}\frac{u}{1-u^2}dx=\int_{-\infty}^{-1}\frac{2u}{1-u^2}dx$$

But I'm sure that I've made some mistake somewhere and can't figure out how to continue further.

Answer 1

Sign mistake here, should be negative:

$$\int_{-\infty}^{-1}\frac{e^x-e^{-x}}{1-e^{-2x}}dx=\int_{-\infty}^{-1}\frac{e^x}{1-e^{-2x}}dx\color {red}{+}\int_{-\infty}^{-1}\frac{e^{-x}}{1-e^{-2x}}dx$$

Here you didnt change the variable dx with $du=e^xdx=udx$

And $u=e^x$ not $u=e^{-x}$ thats different. You confused both...

So $e^{-x}=\dfrac 1 {e^x}= \dfrac 1u$

$$\int_{-\infty}^{-1}\frac{u}{1-u^2}\color{red}{dx}+\int_{-\infty}^{-1}\frac{u}{1-u^2}\color{red}{dx}=\int_{-\infty}^{-1}\frac{2u}{1-u^2}\color{red}{dx}$$

Dont forget to change the bounds ..$ 0 \le u \le \dfrac 1 e$

So here is the correct calculation with your substitution

$$\color{blue}{I=\int_{-\infty}^{-1}\frac{e^x-e^{-x}}{1-e^{-2x}}dx=\int_0^{\dfrac 1 e} \dfrac {u-1/u}{1-1/{u^2}}\frac {du}{u}=\int_0^{\dfrac 1 e}du=\frac 1e}$$

Answer 2

Hint: $$e^{x}-e^{-x} = e^x (1-e^{-2x}).$$

Answer 3

$$\int _{ -\infty }^{ -1 } \frac { e^{ x }-e^{ -x } }{ 1-e^{ -2x } } dx=\int _{ -\infty }^{ -1 } \frac { { e }^{ 2x }\left( { e }^{ 2x }-1 \right) }{ { e }^{ x }\left( { e }^{ 2x }-1 \right) } dx=\int _{ -\infty }^{ -1 }{ { e }^{ x }dx } =\frac { 1 }{ e } $$

Answer 4

observe that

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

the integral becomes

$$\int_{-\infty}^{-1}e^xdx=\Bigl[e^x\Bigr]_{-\infty}^{-1}=\frac {1}{e}-0=1/e$$