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

Question

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

Emma Aldridge · Accepted Answer

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#

Isaiah Allen · Answer

undefined To integrate $ \int \sqrt{x} e^{x-1} \, dx $ using integration by parts:

1. Choose $ u $ and $ dv $.
2. Compute $ du $ and $ v $.
3. Apply the integration by parts formula: $ \int u \, dv = uv - \int v \, du $.

Let's proceed with the integration:

Let $ u = \sqrt{x} $ and $ dv = e^{x-1} \, dx $.
Then, $ du = \frac{1}{2\sqrt{x}} \, dx $ and $ v = e^{x-1} $.

Apply the integration by parts formula:

$$ \int \sqrt{x} e^{x-1} \, dx = uv - \int v \, du $$
$$ = \sqrt{x} \cdot e^{x-1} - \int e^{x-1} \cdot \frac{1}{2\sqrt{x}} \, dx $$

Now, integrate $ \int e^{x-1} \cdot \frac{1}{2\sqrt{x}} \, dx $ using substitution or by recognizing it as a standard integral.

Upon completing the integration, you'll have your final result.