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

2 Answers
Andrea S.
Feb 5, 2017

#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#

Jim H
Feb 5, 2017

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#

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