How do you integrate #dy / (4(y^(1/2))#?

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Answer 1 · 2016-06-03T11:30:29

#1/2y^(1/2)+C#

#[1]" "intdy/(4(y^(1/2)))#

First you can bring #1/4# outside the integral symbol.

#[2]" "=1/4intdy/y^(1/2)#

Next, bring #y^(1/2)# to the numerator.

#[3]" "=1/4inty^(-1/2)dy#

Use the power rule: #intx^ndx=x^(n+1)/(n+1)+C# (where C is a constant)

#[4]" "=1/4*y^(-1/2+1)/(-1/2+1)+C#

#[5]" "=1/4*y^(1/2)/(1/2)+C#

#[6]" "=color(red)(1/2y^(1/2)+C)#