How do you evaluate the integral #int 1/(x+1)^3dx# from -2 to 2? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Andrea S. Nov 30, 2016 #int_(-2)^2 frac 1 ((x+1)^3) dx =-4/9# Explanation: #int_(-2)^2 frac 1 ((x+1)^3) dx = int_(-2)^2 frac 1 ((x+1)^3) d(x+1) =# #= -1/2 frac 1 ((1+x)^2) |_(x=-2)^(x=2) = # #= -1/2 frac 1 ((-2+1)^2) +1/2 frac 1 ((2+1)^2) = -1/2+1/2 * 1/9 = # # =1/2(1/9 -1) = -1/2 * 8/9 =-4/9# 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 1618 views around the world You can reuse this answer Creative Commons License