How do you integrate $int e^(sqrt(2x))$ by parts?

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Answer 1 · 2017-01-19T20:25:26

$inte^sqrt(2x)dx=e^sqrt(2x)(sqrt(2x)-1)+C$

$I=inte^(sqrt(2x))dx$

Let $t=sqrt(2x)$ . This implies that $1/2t^2=x$ , which we differentiate to show that $dx=tcolor(white).dt$ . Then:

$I=inte^t(tcolor(white).dt)=intte^tdt$

We will use integration by parts now, which takes the form $intudv=uv-intvdu$ . For $intte^tdt$ , let:

${(u=t" "=>" "du=dt),(dv=e^tdt" "=>" "v=e^t):}$

Then:

$I=uv-intvdu$

$I=te^t-inte^tdt$

$I=te^t-e^t+C$

$I=e^t(t-1)+C$

Returning to $x$ from $t=sqrt(2x)$ :

$I=e^sqrt(2x)(sqrt(2x)-1)+C$

How do you integrate int e^(sqrt(2x))int e^(sqrt(2x)) by parts?