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

Question

6633 views around the world

You can reuse this answer
Creative Commons License

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#