Question

How do you evaluate the integral `int e^sqrtx`?

Answer 1

`int e^sqrt(x) dx = 2 e^sqrt(x) (sqrt(x)-1) + C`

Explanation:

Substitute:

`t = e^sqrt(x)`

`x = (lnt)^2`

`dx = 2lnt/t dt`

we have:

`int e^sqrt(x) dx = 2 int t lnt/t dt = 2int lnt dt`

we can now integrate by parts:

`int lnt dt = tlnt -int t d(lnt) = tlnt - int t (dt)/t = tlnt -int dt = tlnt -t +C=t(lnt -1) +C`

undoing the substitution:

`int e^sqrt(x) dx = 2 e^sqrt(x) (sqrt(x)-1) + C`

Answer 2

Please see below for an alternative solution.

Explanation:

In order to integrate `e^sqrtx dx` by substitution, we would need the derivative of `sqrtx`. We will introduce the derivative and see if that helps. (If it doesn't help, we'll try something else.)

`int e^sqrtx dx = int underbrace(2sqrtx)_u underbrace(e^sqrtx/(2sqrtx) dx)_(dv)`

With `u = 2sqrtx` and `dv = e^sqrtx/(2sqrtx) dx`, we can find

`du = 1/sqrtx` and (integrating by substitution) `v = e^sqrtx`

`uv-intvdu = 2sqrtx e^sqrtx - int e^sqrtx/sqrtx dx`

`= 2sqrtx e^sqrtx - 2 int e^sqrtx/(2sqrtx) dx`

`= 2sqrtx e^sqrtx - 2e^sqrtx +C`

`= 2e^sqrtx (sqrtx - 1) +C`