If you let u = e^(2x), du = 2e^(2x) dx then you have
1/2 ∫du/(u^2-5u-6)
= 1/2 ∫ du/(u-6)(u+1)
= 1/2 ∫ 1/7 (1/(u-6) - 1/(u+1)) du
so, yeah, that looks good to me.
wolframalpha went one step further, using tanh-1x = 1/2 ln |(1+x)/(1-x)| to wind up with
1/7 tanh-1[(5 - 2e^(2x))/7]
∫ (e^2x/e^4x-5e^2x-6)dx
Is this right
1/14 ln |e^2x-6| - 1/14 ln |e^2x+1| +c
1 answer