How do you integrate #(5/sqrt x) dx#?

1 Answer
Aug 4, 2016

#int 5/(sqrt(x))d x=10 sqrt x +C#

Explanation:

#int 5/(sqrt(x))d x=?#

#"we can rearrange the term as;" #

#int 5(x)^(-1/2)d x=5 int(x)^(-1/2) d x#

#int int 5/(sqrt(x))d x=5(1/(-1/2+1)) x^(-1/2+1)#

#int 5/(sqrt(x))d x=5*2*x^(1/2)#

#int 5/(sqrt(x))d x=10 sqrt x +C#