weird integral?

Question

how would you take the integral of S ((2e^x - 2e^(-x))/(e^x+e^(-x))^2 dx?
i did e^x+e^(-x) as the u, which is at the top if u factor out 2, but then im left with 2 S (1/U^2) du would i just take it as a regular integral? like what i got as the answer is 2ln| (e^x - e ^-x )^2 | +C?
is this wrong?

even if the denom. is U^2? 1/U^2 du = ln | U^2| + C? is that right what i did?

ooh haha i didnt see that. i was so focused on finding another ln in there. thanks!!!

Answer 1

∫[((2e^x - 2e^(-x))/(e^x + e^(-x))^2]dx =
2∫[((e^x - e^(-x))/(e^x + e^(-x))^2]dx =
let u = e^x + e^(-x), du = (e^x - e^(-x))dx
2∫[((e^x - e^(-x))/(e^x + e^(-x))^2]dx =
2∫du/u^2 =
-2u^-1 + C =
-2/(e^x + e^(-x)) + C

Answer 2

You have mismatched parentheses. I assume you mean:
Integrate ∫{[2e^x - 2e^(-x)]/[e^x+e^(-x)]}² dx.

First let's work algebraically with the expression to make it easier to integrate.

{[2e^x - 2e^(-x)]/[e^x+e^(-x)]}² = 2tanh²x = 2sinh²x/cosh²x
= 2(cosh²x - 1)/cosh²x = 2(1 - sech²x)

Now we can integrate.

∫{[2e^x - 2e^(-x)]/[e^x+e^(-x)]}² dx
= 2∫(1 - sech²x)dx = 2[x - tanh(x)] + C

Answer 3

No U are tight. U know inttegral of U'/U is Ln(U).

Answer 4

Now that is some crazy s***. Answer is -sech(x) + C.