Integrate $\int_{-\infty}^{-1}\frac{e^x-e^{-x}}{1-e^{-2x}}dx$ 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.
 A: Hint: $$e^{x}-e^{-x} = e^x (1-e^{-2x}).$$
A: observe that
$$e^x-e^{-x}=e^x (1-e^{-2x} )$$
the integral becomes
$$\int_{-\infty}^{-1}e^xdx=\Bigl[e^x\Bigr]_{-\infty}^{-1}=\frac {1}{e}-0=1/e$$
A: 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
$$\color{blue}{I=\int_{-\infty}^{-1}\frac{e^x-e^{-x}}{1-e^{-2x}}dx=\int_0^{\dfrac 1 e} \dfrac {u-1/u}{1-1/{u^2}}\frac {du}{u}=\int_0^{\dfrac 1 e}du=\frac 1e}$$
A: $$\int _{ -\infty  }^{ -1 } \frac { e^{ x }-e^{ -x } }{ 1-e^{ -2x } } dx=\int _{ -\infty  }^{ -1 } \frac { { e }^{ 2x }\left( { e }^{ 2x }-1 \right)  }{ { e }^{ x }\left( { e }^{ 2x }-1 \right)  } dx=\int _{ -\infty  }^{ -1 }{ { e }^{ x }dx } =\frac { 1 }{ e } $$
