How do you evaluate the definite integral #int (1/x^2-1/x^3)dx# from [1,2]? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Ratnaker Mehta Jan 28, 2017 #1/8#. Explanation: Since, #intx^ndx=x^(n+1)/(n+1)+C, n!=-1#, we have, #int_1^2(1/x^2-1/x^3)dx=[x^(-2+1)/(-2+1)-x^(-3+1)/(-3+1)]_1^2# #=[x^-1/-1-(x^-2/-2)]_1^2=[1/(2x^2)-1/x]_1^2=(1/8-1/2)-(1/2-1)# #=-3/8-(-1/2)=1/8#. 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 9962 views around the world You can reuse this answer Creative Commons License