Mathematics OpenStudy (anonymous):

Integral for fun :) OpenStudy (anonymous):

$\int\limits \frac{e^{3x}}{(1+e^x)^3}dx$ OpenStudy (anonymous):

And this one: $\int\limits_{-\infty}^{\infty}\frac{x^2-x-1}{x^4+10x^2+9}dx$ OpenStudy (anonymous):

$\int\limits_{}^{}e ^{2x}*e ^{x}dx/(1+e ^{x})^{3}$ 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 OpenStudy (anonymous):

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 OpenStudy (anonymous):

At what point did you make that substitution? At the beginning? Or after parts? OpenStudy (anonymous):

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. OpenStudy (anonymous):

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 :) OpenStudy (anonymous):

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

Tomorrow though, I'm about to hit the bed :P OpenStudy (anonymous):

Haha, I gotcha. Yeah, I've never done complex analysis. Sounds crazy, in a good way. OpenStudy (anonymous):

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

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