Question

Integration
\[\int \sqrt{1+e^x}dx\]

Answer 1

Let \( u =\sqrt{1+e^x}\)
\[du = \frac{e^x}{2\sqrt{1+e^x}}dx\]\[dx = \frac{2\sqrt{1+e^x}}{e^x}du = \frac{2u}{(u^2-1)}du\]
Then the integral becomes
\[\int u (\frac{2u}{(u^2-1)}du)\]Am I doing anything wrong here?

Answer 2

That's right now just write u^2 as u^2-1 +1

Answer 3

nopes this is absolutely correct continue..

Answer 4

after that you will get only 1/(u^2 -1) which can be written as 1/(u-1)(u+1) which can be further written as \[\frac{ (u+1) -(u-1) }{ (u+1)(u-1) }\] = \[\frac{ 1 }{(u-1) } - \frac{ 1 }{u+1 }\]

Answer 5

now you integrate easily did you get it rolypoly

Answer 6

Hold on, just a second, I'm working on it..

Answer 7

How come '' you will get only 1/(u^2 -1)''??

Answer 8

\[\int u (\frac{2u}{(u^2-1)}du)\]\[=2\int \frac{u^2}{u^2-1}du\]\[=2\int 1+\frac{1}{u^2-1}du\]\[=2u+2\int \frac{1}{u^2-1}du\]\[=2u+\int \frac{1}{u-1}-\frac{1}{u+1}du\]\[=2u+\ln |u-1|-\ln |u+1| +C\]\[=2\sqrt{1+e^x}+\ln |\sqrt{1+e^x}-1|-\ln |\sqrt{1+e^x}+1| +C\]Anything wrong here?

Answer 9

no this is correct nicely done rolypoly

Answer 10

Thanks a lot!! :)