How do you integrate #int sqrtx e^(x-1 ) dx # using integration by parts?

1 Answer
May 22, 2016

Please see the explanation section below.

Explanation:

First note that #int sqrtx e^(x-1 )dx = 1/e int sqrtxe^x dx #

Let #t = sqrtx#, so that #dt = 1/(2sqrtx) dx# and, therefore #dx = 2t#

Upon substitution, the integral becomes:

#1/e int te^(t^2) (2t) dt#

Now let #u = t# and #dv = e^(t^2) (2t) dt#

so that we get #du = dt# and #v = e^(t^2)# (integrate by substitution.

Apply the parts rule to get

#1/e [te^(t^2) - int e^(t^2) dt]#

#int e^(t^2) dt = sqrtpi/2 "erfi"(t)# (the imaginary error function at #t#), so we get

#1/e [te^(t^2) - sqrtpi/2 "erfi"(t)] +C#

Reversing our substitution gets us:

#int sqrtx e^(x-1 )dx = 1/e[sqrtxe^sqrtx - sqrtpi/2 "erfi"(sqrtx)]+C#