exam question

**^_^Engineer_Adam^_^** · June 28th 2007, 06:18 AM

this is my minus 20 points
feel free to answer:
$\int\frac{e^{2x}+4}{e^{2x}-4}dx$

**galactus** · June 28th 2007, 06:34 AM

You could have written it as:

$\frac{2}{e^{x}-2}-\frac{2}{e^{x}+2}+1$

and went from there.

$\int\frac{2}{e^{x}-2}dx=[ln(e^{x}-2)-x]$

$\int\frac{2}{e^{x}+2}dx=[x-ln(e^{x}+2)]$

**red_dog** · June 28th 2007, 06:37 AM

$\displaystyle \int\frac{e^{2x}+4}{e^{2x}-4}dx=\int\left(1+\frac{8}{e^{2x}-4}\right)dx=x+8\int\frac{1}{e^{2x}-4}dx$ .
Substitute: $\displaystyle e^{2x}=t\Rightarrow 2x=\ln t\Rightarrow dx=\frac{1}{2t}dt$ .
Then $\displaystyle 4\int\frac{dt}{t(t^2-4)}=-\ln |t|+\frac{1}{2}\ln |t+2|+\frac{1}{2}\ln |t-2|+C$ .
So the original integral is
$\displaystyle -x+\frac{1}{2}\ln |e^{2x}-2|+\frac{1}{2}\ln (e^{2x}+2)+C$

**Soroban** · June 28th 2007, 07:05 AM

Hello, ^_^Engineer_Adam^_^!

Another approach . . .

$\int\frac{e^{2x}+4}{e^{2x}-4}dx$

Let $e^{2x} = u\quad\Rightarrow\quad x = \frac{1}{2}\ln u\quad\Rightarrow\quad dx = \frac{du}{2u}$

Substitute: . $\int\frac{u+4}{u-4}\left(\frac{du}{2u}\right) \;=\;\frac{1}{2}\int\frac{u+4}{u(u-4)}\,du$

Partial Fractions: . $\frac{1}{2}\int\left(\frac{2}{u-4} - \frac{1}{u}\right)du\;=\;\frac{1}{2}\left[2\ln(u-4) - \ln u\right] + C<br />$

Back-substitute: . $\frac{1}{2}\left[2\ln(e^{2x} - 4) - \ln(e^{2x})\right] + C \;=\;\frac{1}{2}\left[2\ln(e^{2x} - 4) - 2x\right] + C$

. . . . . . . . . . $= \;\ln(e^{2x}-4) - x + C$

**^_^Engineer_Adam^_^** · July 1st 2007, 03:19 AM

or in the integral $\int \frac{4}{e^{2x}-4}dx$ you could multiply both numerator and denominator by
$\frac{e^{-2x}}{e^{-2x}}$

btw thanks for the help!

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