How do you integrate int sqrtx e^x dx using integration by parts?

1 Answer
Jun 10, 2018

sqrtxe^x-(sqrtpierfi(sqrtx))/2+C

Explanation:

Problem:
intsqrtxe^xdx

Integrate by parts: intfgprime=fg-intfprimeg
f=sqrtx, gprime=e^x
darr
fprime=1/(2sqrtx),g=e^x:
=sqrtxe^x-inte^x/(2sqrtx)dx

Now solving:
inte^x/(2sqrtx)dx
Substitute u=sqrtx->dx=2sqrtxdu:
=sqrtpi/2int(2e^[u^2])/sqrtpidu

Now solving:
int(2e^[u^2])/sqrtpidu

This is a special integral, imaginary error function:
=erfi(u)

Plug in solved integrals:
sqrtpi/2int(2e^[u^2])/sqrtpidu
=[sqrtpierfi(u)]/2

Undo substitution u=sqrtx:
=[sqrtpierfi(sqrtx)]/2

Plug in solved integrals:
sqrtxe^x-inte^x/2sqrtxdx
=sqrtxe^x-[sqrtpierfi(sqrtx)]/2

The problem is solved:
intsqrtxe^xdx
=sqrtxe^x-(sqrtpierfi(sqrtx))/2+C