How do you integrate #(5x)/ (x-2)^(1/2) dx#?

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Answer 1 · 2015-04-11T15:04:20

#f(x)=5intx/(x-2)^(1/2)dx#

Let's #u = x-2#

#du = 1#

#x = u +2#

Now we have : #=5int(u+2)/u^(1/2) du#

#=5intsqrt(u)du+10int1/sqrt(u)du#

#=[10/3sqrt(u^3)]+20[sqrt(u)]+C#

Substitute back

#10/3(x-2)^(3/2)+20(x-2)^(1/2)+C#