How do you evaluate the integral #int 1/(x-2)^(2/3)dx# from 1 to 4? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Andrea S. Dec 30, 2016 #int_1^4 (dx)/(x-2)^(2/3) = 3(root(3)2+1)# Explanation: #int_1^4 (dx)/(x-2)^(2/3) = [3(x-2)^(1/3)]_1^4 = 3(2^(1/3)-(-1)^(1/3))= 3(root(3)2+1)# 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 4553 views around the world You can reuse this answer Creative Commons License