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#