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

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Answer 1 · 2016-05-22T13:47:26

Please see the explanation section below.

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:

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

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