How do you evaluate the integral #int 1/x^3dx# from -1 to 2? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Alan N. Aug 23, 2016 #3/8# Explanation: #int_-1^2 1/x^3 dx = int_-1^2 x^-3 dx# #= x^-2/-2|_-1 ^2# #= -1/(2x^2)|_-1 ^2# #=-1/8- (-1/2)# #=3/8# 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 3867 views around the world You can reuse this answer Creative Commons License