Question #6b2a6 Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Cesareo R. Aug 23, 2016 #sqrt(2y-3y^2) +C# Explanation: #d/(dy)sqrt(2y-3y^2) = 1/2(2-6y)/sqrt(2y-3y^2) =(1-3y)/sqrt(2y-3y^2) # so #int (1-3y)/(sqrt(2y-3y^2))dy = sqrt(2y-3y^2) +C# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 1097 views around the world You can reuse this answer Creative Commons License