How do you evaluate the definite integral #int (x^2-x-2)/(x-2)dx# from [0,1]? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Ratnaker Mehta Jul 12, 2017 # 3/2.# Explanation: #I=int_0^1(x^2-x-2)/(x-2)dx=int_0^1{(x-2)(x+1)}/(x-2)dx.# Since, #x in [0,1], x!=2, :.," so, dividing by "(x-2)ne0,# we have, #I=int_0^1(x+1)dx,# #=[x^2/2+x]_0^1.# # rArr I=3/2.# 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 1244 views around the world You can reuse this answer Creative Commons License