How do you evaluate #int (x^2-1)/( x+1) dx# for [1, 2]? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Dharma R. Sep 19, 2015 #1/2# Explanation: #int_1^2(x^2-1)/(x+1)dx=int_1^2((x+1)(x-1))/(x+1)dx# =#int_1^2(x-1)dx=x^2/2-x|_1^2# =#2^2/2-2-(1/2-1)# =#1/2# 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 2516 views around the world You can reuse this answer Creative Commons License