Question

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

Answer 1

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`