How do you integrate #dy / (4(y^(1/2))#? Calculus Introduction to Integration Integrals of Polynomial functions 1 Answer Rafael Jun 3, 2016 #1/2y^(1/2)+C# Explanation: #[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)# Answer link Related questions How do you evaluate the integral #intx^3+4x^2+5 dx#? How do you evaluate the integral #int(1+x)^2 dx#? How do you evaluate the integral #int8x+3 dx#? How do you evaluate the integral #intx^10-6x^5+2x^3 dx#? What is the integral of a constant? What is the antiderivative of the distance function? What is the integral of #|x|#? What is the integral of #3x#? What is the integral of #4x^3#? What is the integral of #sqrt(1-x^2)#? See all questions in Integrals of Polynomial functions Impact of this question 2195 views around the world You can reuse this answer Creative Commons License