How do you find the indefinite integral of #int (5sqrty-3/sqrty)dy#?

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Answer 1 · 2016-08-31T14:49:21

#2/3y^(1/2)(5y-9)+C#,

or,

#2/3sqrty(5y-9)+C#.

Suppose that, #I=int(5sqrty-3/sqrty)dy=5inty^(1/2)dy-3inty^-(1/2)dy#.

Since. #intx^n dx=x^(n+1)/(n+1), n!=-1#, we have,

#I=5y^(1/2+1)/(1+1/2)-3y^(-1/2+1)/(-1/2+1)#

#=10/3y^(3/2)-6y^(1/2)#.

#=2/3y^(1/2)(5y-9)+C#, or,

#=2/3sqrty(5y-9)+C#.