Question

I need some help with the homework problem. I have a list of formulas, but can't seem to get it done.


`\int\frac{9}{\sqrt{1+e^(2x)}} \, dx`

I started by taking the constant out and setting u =
`\sqrt{1+e^(2x\\)}`
After this I can't seem to progress.

Answer 1

After setting
`u=\sqrt{1+e^(2x)}`, partially solving for x in terms of u gives

`u = sqrt{1+e^(2x)} \implies u^2 = 1 + e^(2x) \implies e^(2x) = u^2 - 1`

Then taking differentials, you get

`2 e^(2x) \,\mathrm dx = 2u \, \mathrm du \implies \mathrm dx = (u)/(u^2-1)\,\mathrm du`

Replacing everything in the original integral then gives

`\displaystyle \int \frac9{\sqrt{1+e^(2x)}}\,\mathrm dx = \int \frac9u * \frac u{u^2-1}\,\mathrm du = 9 \int (\mathrm du)/(u^2-1)`

Split up the integrand into partial fractions:

`\frac1{u^2-1} = \frac a{u-1} + \frac b{u+1} \\\\ 1 = a(u+1) + b(u-1) = (a+b)u + a-b \\\\ \implies \begin{cases}a+b=0\\a-b=1\end{cases} \implies a=\frac12,b=-\frac12`

so that

`\displaystyle 9 \int (\mathrm du)/(u^2-1) = \frac92 \int \left(\frac1{u-1} - \frac1{u+1}\right) \,\mathrm du \\\\ = \frac92 \left(\ln|u-1| - \ln|u+1|\right) + C \\\\ = \frac92 \ln\left|(u-1)/(u+1)\right| + C \\\\ = \frac92 \ln\left(\frac{\sqrt{1+e^(2x)}-1}{\sqrt{1+e^(2x)}+1}\right) + C`