How do you integrate #int (x^2-1)/( x+1 )dx#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Lovecraft Nov 14, 2015 #int((x^2-1)/(x+1)) = (x^2)/2 -x+c# Explanation: We know that #x^2 - 1 = (x+1)(x-1)#, by the difference of squares so we can rewrite the integral to be #int((x+1)(x-1))/(x+1)dx = int(x-1)dx = intxdx - intdx = x^2/2 - x + c# Or if you prefer #int((x^2-1)/(x+1)) = (x^2-2x+c)/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 27032 views around the world You can reuse this answer Creative Commons License