-1
$\begingroup$

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

  • 2
    Did you try $u=e^x$? – 2017-02-15

3 Answers 3

1

Hint. Use the substitution $y=e^x$.

If you really want to do it with the notions within the section, you could rewrite the integrand as follows:

$$\int\frac{e^x}{e^{2x}-1} dx = \int \frac{2e^{-x}}{2e^{-x}} \frac{e^x}{e^{2x}-1}\ dx = \frac{1}{2}\int\frac{2}{e^x - e^{-x}}dx = \frac{1}{2}\int\frac{1}{\sinh(x)}dx$$ which does not really simplify matters.

  • 0
    I didn't think it was that easy because the section I am on involves sinh, cosh, etc and their inverses. – 2017-02-15
  • 0
    It was that easy, check my answer – 2017-02-15
  • 0
    @someonee See my updated answer. – 2017-02-15
1

Based on the background, you are looking at

$$\int\frac{e^x}{e^{2x}-1}\ dx=\int\frac1{e^x-e^{-x}}\ dx=\frac12\int\operatorname{csch}(x)\ dx$$

Intuition says

$$\int\operatorname{csch}(x)\ dx=\ln(\tanh\frac x2)+c$$

since,

$$\int\csc(x)\ dx=\ln(\tan\frac x2)+c$$

Two hints:

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

Somehow, you'll want to end up with

$$\int\frac1{u^2-1}\ du$$

And PFD reveals

$$\frac1{u^2-1}=\frac12\left(\frac1{u-1}-\frac1{u+1}\right)$$

  • 1
    Your partial fraction is wrong it's $\frac{1}{2}\left (\frac{1}{u-1}\color{red}-\frac{1}{u+1}\right )$ – 2017-02-15
  • 0
    Oops, probably was a typo. My bad! – 2017-02-15
0

Just use substitution, where

$u=e^x$

$\int \frac{u}{u^2-1}\cdot \frac{1}{u}du$, this equals

$=\int \frac{1}{-\left(-u^2+1\right)}du$

$=-arctanh \left(e^x\right)+C$

  • 2
    Didn't down-vote but It's $-1$ not $+1$. – 2017-02-15
  • 1
    Damn, gonna edit, thanks. – 2017-02-15
  • 1
    You may wish to check that. – 2017-02-15
  • 0
    @SimplyBeautifulArt Yeah silly mistakes that come from copying from my notebook, thanks. – 2017-02-15