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$\int e^x·\sqrt{1+e^x}\,dx$

Can someone explain to me how to work this integral out?
I tried integration by parts, however, that does not give a clear answer.
I got a hint from our professor that we should use substitution, but if I substitute "$ 1+e^x$" for example, I still can't calculate it by integration by parts...

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    $\begingroup$ let $1+e^x=u$. . $\endgroup$
    – Nosrati
    Nov 26, 2017 at 14:42

2 Answers 2

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Substitute $1+e^{ x }=u$ then $du={ e }^{ x }dx$

$$\\ \\ \int e^{ x }\sqrt { 1+e^{ x } } \,dx=\int \sqrt { u } \,du=\frac { 2u\sqrt { u } }{ 3 } +C=\\ =\frac { 2 }{ 3 } { \left( { e }^{ x }+1 \right) }^{ \frac { 3 }{ 2 } }+C$$

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  • $\begingroup$ It's $\frac23(1+e^x)^\frac32+C$ not $\frac32(e^x)^\frac32+C$. $\endgroup$
    – TheSimpliFire
    Nov 26, 2017 at 14:47
  • $\begingroup$ @TheSimpliFire,it was a typo,thank you $\endgroup$
    – haqnatural
    Nov 26, 2017 at 14:53
  • $\begingroup$ Note also the inversion of $\frac32$ :) $\endgroup$
    – TheSimpliFire
    Nov 26, 2017 at 14:55
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    $\begingroup$ today is not my day ( $\endgroup$
    – haqnatural
    Nov 26, 2017 at 15:00
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    $\begingroup$ You forgot a : in your smily LOL $\endgroup$
    – user223391
    Nov 26, 2017 at 15:03
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Do the substitution $1+e^x=t$ and $e^x\,\mathrm dx=\mathrm dt$.

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