roadjester
  • roadjester
integrate x*e^x/sqrt(1+e^x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Make the substitution that u=sqrt(1+e^x) and you should be able to integrate it :)
anonymous
  • anonymous
oh wait sorry I didn't see the x in front there
roadjester
  • roadjester
\[\int\limits_{?}^{?} {xe ^{x}\over \sqrt{1+e^x}}dx\]]

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anonymous
  • anonymous
integrate by parts with u = x and dv/dx = e^x/sqrt(1+e^x) you then do the above substitution to find v
roadjester
  • roadjester
@eigenschmeigen Final answer: \[2(x-2)\sqrt{1+e ^{x}}+2{\ln {\sqrt{{1+e ^{x}}}+1\over \sqrt{1+e ^{x}}-1}}+C\] When I did the substitution you suggested, I didn't get the answer in the book.
anonymous
  • anonymous
\[\int\limits{\frac{e^x}{\sqrt{1+e^x}}}dx \] let \[u = 1+e^x\] \[\frac{du}{dx} = e^x\] \[dx= \frac{du}{e^x}\] \[\int\limits\limits{\frac{e^x}{\sqrt{1+e^x}}}dx = \int\limits\limits{\frac{e^x}{\sqrt{u}}}\frac{du}{e^x}\]\[=\int\limits\limits\limits{\frac{1}{\sqrt{u}}}du\]
anonymous
  • anonymous
i dont really have time to finish, sorry
roadjester
  • roadjester
@eigenschmeigen That doesn't make any sense! That integral would give me:\[2\sqrt{1+e ^{x}}+C\]
myininaya
  • myininaya
This was his plan for you: \[\int\limits_{}^{} x \cdot \frac{e^x}{\sqrt{1+e^x}} dx= x \cdot 2 \sqrt{1+e^x}-\int\limits_{}^{} (x)' 2 \sqrt{1+e^x} dx\]
myininaya
  • myininaya
myininaya
  • myininaya
Ok I will be back on tomorrow if more help is needed peace
roadjester
  • roadjester
@myininaya I'm not sure where your method progresses to, but an integration by parts plus a trig substitution is too complex. No offense.
apoorvk
  • apoorvk
I think roadjester's method is a bit easier, but then, myininaya's way is another new thing I learnt today. :)
apoorvk
  • apoorvk
Sorry, I meant eigenschmeigen's instead of roadjester's.
dumbcow
  • dumbcow
http://www.wolframalpha.com/input/?i=integrate+%28xe^x%29%2Fsqrt%281%2Be^x%29+dx
anonymous
  • anonymous
yeah, by parts like @myininaya has done, then it requires a trig substitution

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