How do you find the most general antiderivative or indefinite integral #int (1/7-1/y^(5/4)) dy#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Ratnaker Mehta Feb 7, 2017 #1/7y-5/y^(1/4)+C.# Explanation: Recall that, #(1): inty^ndy=y^(n+1)/(n+1)+c, n!=-1.# #(2): int[kf(y)-g(y)]dy=kintf(y)dy-intg(y)dy, k" is a const."# #"Therefore, "int(1/7-1/y^(5/4))dy# #=1/7inty^0dy-inty^(-5/4)dy# #=1/7y^(0+1)/(0+1)-(-5/4)(y^(-5/4+1))/((-5/4+1))# #=1/7y+{(5/4)/(-1/4)}y^(-1/4)# #=1/7y-5/y^(1/4)+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 2421 views around the world You can reuse this answer Creative Commons License