Recent content by Celadon

Thank you very much for all your help! I can finally move on and not worry about this one question anymore. The sense of relief from when you solve a math problem.. words can't describe it. I really do appreciate your help. :D

Just an emoticon; I add them to a lot of my sentences out of habit, haha... Sorry. Okay so, I tried it again like this: u=ex du=exdx dx=du(1/ex) ∫u2dx-2x+∫u-2dx = ∫u2(1/u)du-2x+∫u-2(1/u)du =∫udu - 2x + ∫u-3du =u2/2 - 2x + (u-2/-2) + C =u2/2 - 2x - 1/(2u2) + C =e2x/2 - 2x - 1/(2e2x) + C

(u2)/2? :x

u1 or just u? xD (also, I edited my previous post with another answer in case you missed it.)

∫u2dx? :x [EDIT] Another attempt: 2e2x-2x-2e-2x+C x_x

ex+C?

Okay, another attempt. Hope I fixed more than I broke, this time. :/ Thanks again everyone, I'm sorry for bothering you all. D: ∫(ex-e-x)2dx = ∫(e2x-2+e-2x)dx = ∫e2xdx + ∫e-2xdx - ∫2dx Let u = ex du = exdx ∫e2xdx + ∫e-2xdx - ∫2dx = (1/ex)(u3/3)+(1/ex)(u-1/-1)-2x+C =...

Oh I see it now. (ex - e-x)(ex - e-x) = e2x - 2 + e-2x I somehow wrote "+ e0 - e0" at some point, oops. Thanks! Time to try again. (:

Okay, do I tried expanding it and then went through this: (ex-e-x)2 = (ex-e-x)(ex-e-x) = e2x + e-2x ∫(e2x + e-2x) dx u = e2x du = (e2x / 2) dx Final solution: (u2/e2x) + (2 / e2x)ln|e2x|

Before I continue asking questions, thanks for all the help! I really do appreciate it and I'm sorry if I'm a bit slow, haha. Now then... @BruceW: Thanks for the help, the welcome, and the name compliment! I'm actually a fan of the colour itself and the city in Pokemon, heh. @rock.freak667...

Homework Statement Find the indefinite integral (preferably using u-substitution): ∫(ex-e-x)2 dx Homework Equations N/A The Attempt at a Solution To be honest, I'm slightly confused as to which path I'm supposed to take with this, especially since I'm not sure what I should be...