integrate Sqrt(1+e^x) ?
multiply both top and bottom by e^x, then let u = e^x
do that and what do you get?
im not sure!
i dont understand how that helps!
sqrt(1+e^x) e^x dx / e^x u = e^x, du = e^x dx sqrt(1 + u)/u du yes?
yeah that makes sense
now let t = sqrt(1 + u), what do you get?
t/u du?
solve for u
I'm sorry after spring break i feel so dumb, i'm not sure
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is this some what on the right path?
you could also try making \(u=1+e^x\)
t = sqrt(1 + u) t^2 = 1 + u u = t^2 - 1 du = 2t dt t (2t) / (t^2 - 1) dt 2t^2 / (t^2 - 1) The integral now turns to partial composition type
another way to skin the cat, go the whole hog and make \[y=\sqrt{1+e^x}\]
you get \[u=\sqrt{1+e^x}\] \[u^2=1+e^x\] \[u^2-1=e^x\] \[\ln(u^2-1)=x\] \[dx=\frac{1}{u-1}+\frac{1}{u+1}\]
so to get from 2t^2/t^2-1 dt to the last step you posted I just use partial fractions? @sourwing
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