How do you find the integral of #f(x)=xsqrt(x-5)# using integration by parts?

1 Answer
Nov 16, 2015

#2/3 x (x-5)^(3/2) - 4/15 (x-5)^(5/2)#

Explanation:

First, here's the formula for "integration by parts":

#int f'(x) * g(x) "d"x = f(x) g(x) - int f(x) g'(x) "d" x #

Now, in your product #x * sqrt(x-5)# you need to determine which factor is #f'(x)# and which one is #g(x)# since you will need to integrate one of them and differentiate the other one.

As it doesn't make a big difference with the radical expression but would be a good thing to differentiate #x#, let's assume that #f(x) = x# and #g'(x) = sqrt(x-5)#.

As next, you need to differentiate #f(x)# and integrate #g'(x)#:
#f(x) = x => f'(x) = 1#
#g'(x) = sqrt(x-5) = (x-5) ^(1/2) => g(x) = 2/3 * (x-5)^(3/2)#

Now we can use the formula:

#int x * sqrt(x-5) "d"x#
#= x * 2/3 * (x-5)^(3/2) - int 1 * 2/3 * (x-5)^(3/2) "d"x#
# = 2/3 x (x-5)^(3/2) - 2/3 int (x-5)^(3/2) "d"x#
# = 2/3 x (x-5)^(3/2) - 2/3 *2/5 (x-5)^(5/2)#
# = 2/3 x (x-5)^(3/2) - 4/15 (x-5)^(5/2)#

Hope that this helped!