Ask your own question, for FREE!
Calculus1 12 Online
OpenStudy (anonymous):

integrate Sqrt(1+e^x) ?

OpenStudy (anonymous):

multiply both top and bottom by e^x, then let u = e^x

OpenStudy (anonymous):

do that and what do you get?

OpenStudy (anonymous):

im not sure!

OpenStudy (anonymous):

i dont understand how that helps!

OpenStudy (anonymous):

sqrt(1+e^x) e^x dx / e^x u = e^x, du = e^x dx sqrt(1 + u)/u du yes?

OpenStudy (anonymous):

yeah that makes sense

OpenStudy (anonymous):

now let t = sqrt(1 + u), what do you get?

OpenStudy (anonymous):

t/u du?

OpenStudy (anonymous):

solve for u

OpenStudy (anonymous):

I'm sorry after spring break i feel so dumb, i'm not sure

OpenStudy (anonymous):

|dw:1395892978591:dw|

OpenStudy (anonymous):

is this some what on the right path?

OpenStudy (anonymous):

you could also try making \(u=1+e^x\)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

another way to skin the cat, go the whole hog and make \[y=\sqrt{1+e^x}\]

OpenStudy (anonymous):

you get \[u=\sqrt{1+e^x}\] \[u^2=1+e^x\] \[u^2-1=e^x\] \[\ln(u^2-1)=x\] \[dx=\frac{1}{u-1}+\frac{1}{u+1}\]

OpenStudy (anonymous):

so to get from 2t^2/t^2-1 dt to the last step you posted I just use partial fractions? @sourwing

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!