How do you integrate $int e^sqrtx$ by integration by parts method?

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Answer 1 · 2016-09-25T18:53:08

$2e^sqrtx(sqrtx-1)+C$

$inte^sqrtxdx$

Let $t=sqrtx$ . This also implies that $t^2=x$ , and this makes finding $dx$ easier, in that we easily see that $2tdt=dx$ . Thus:

$inte^sqrtxdx=int2te^tdt=2intte^tdt$

Now, using integration by parts, which takes the form $intudv=uv-intvdu$ . Let:

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

Thus:

$2intte^tdt=2[te^t-inte^tdt]=2te^t-2e^t+C$

Factoring and back-substituting with $t=sqrtx$ :

$inte^sqrtxdx=2e^sqrtx(sqrtx-1)+C$

How do you integrate int e^sqrtxint e^sqrtx by integration by parts method?