Question

Integral for fun :)

Answer 1

\[\int\limits \frac{e^{3x}}{(1+e^x)^3}dx\]

Answer 2

And this one:
\[\int\limits_{-\infty}^{\infty}\frac{x^2-x-1}{x^4+10x^2+9}dx\]

Answer 3

\[\int\limits_{}^{}e ^{2x}*e ^{x}dx/(1+e ^{x})^{3}\]

I solved this on paper. A bit too lazy to repost it here, but basically after rewriting it as above, we use integration by parts with u=e^(2x) and dv=e^x/(1+e^x)^3....the dv part is easy to integrate with a basic u-substitution, a la u=1+e^x. Once we integrate everything, we end up with a second integrand, e^2x/(1+e^x)^2. This is a tad trickier, but basically we rewrite e^2x as e^x * e^x, then use a u-substitution with u=1+e^x, then e^x=u-1 and du=e^(x)dx, which gives us an integrand of (u-1)/u^2, which is pretty easy to integrate. The final answer is

(-1/2)e^(2x)/(1+e^x)^2+ln(abs(1+e^x))+1/(1+e^x)+C

Answer 4

Damn that's alot xP Let u=e^x
then du=e^xdx
or
1/e^x du=dx
but e^x=u
so
1/u du=dx
plug it in and do partial fractions :P
What about the second one???? :P

Answer 5

At what point did you make that substitution? At the beginning? Or after parts?

Answer 6

The very beginning. No parts required. Then the integral becomes:
\[\int\limits \frac{u^3}{u(1+u)^3}du\]
Then a straight forward partial fractions.

Answer 7

Ah, very nice. Oh well, I hate partial fractions anyway :)

I'll see what I can do on the second one, though I kind of hate improper integrals too :)

Answer 8

Thats why complex analysis is fantastic :P If you want I can explain residue theory for you xP

Answer 9

Tomorrow though, I'm about to hit the bed :P

Answer 10

Haha, I gotcha. Yeah, I've never done complex analysis. Sounds crazy, in a good way.

Answer 11

It is :D But if you want, I can explain how to make the above problem easy and fairly quick using complex analysis. :D Night xP