How do you find the integral of #int 1/ (x-2)^2 dx # from 0 to infinity? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Dharma R. Sep 20, 2015 -1 Explanation: #int_0^oo1/(x-2)^2dx=int_0^oo(x-2)^(-2)dx=[(x-1)^(-1)/(-1)]_0^oo# =#[1/(x-1)]_oo^0=1/(0-1)-1/(oo-1)=-1-1/oo=-1# 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 3110 views around the world You can reuse this answer Creative Commons License