Question

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

Answer 1

`intxe^sqrt(x)dx = 2e^sqrt(x)(x^(3/2)-3x+6sqrt(x)-6)+C`

Explanation:

First, we use substitution.

Let `t = sqrt(x) => dt = 1/(2sqrt(x))dx` and `t^3 = x^(3/2)`

Then, substituting, we have

`intxe^sqrt(x)dx = 2intx^(3/2)e^sqrt(x)1/(2sqrt(x))dx = 2intt^3e^tdt`

Next, we apply integration by parts three times, using the formula

`intudv = uv - intvdu`

Integration by Parts 1:

Let `u = t^3` and `dv = e^tdt`
Then `du = 3t^2dt` and `v = e^t`

Applying the formula:
`2intt^3e^tdt = 2(t^3e^t - 3intt^2e^tdt)`

Integration by Parts 2:

Focusing on the remaining integral, let `u = t^2` and `dv = e^tdt`
Then `du = 2tdt` and `v = e^t`

Applying the formula:
`intt^2e^tdt = t^2e^t - 2intte^tdt`

Integration by Parts 3:

Again, focusing on the remaining integral, let `u = t` and `dv=e^tdt`
Then `du = dt` and `v = e^t`

Applying the formula:
`intte^tdt = te^t-inte^tdt = te^t - e^t + C`

Substituting our result into the second integration by parts step, we obtain

`intt^2e^tdt = t^2e^t-2(te^t-e^t) + C = e^t(t^2-2t+2)+C`

Substituting this into the first integration by parts step, we obtain

`2intt^3e^tdt = 2[t^3e^t-3e^t(t^2-2t+2)]+C`

`=2e^t(t^3-3t^2+6t-6)+C`

Finally, we substitute `t=sqrt(x)` back in to obtain our final result:

`intxe^sqrt(x)dx = 2e^sqrt(x)(x^(3/2)-3x+6sqrt(x)-6)+C`