What is the antiderivative of #(1/(x^2))#? Calculus Introduction to Integration Integrals of Polynomial functions 1 Answer Jim H Apr 18, 2016 #-1/x+C# Explanation: #d/dx(x^r+C) = rx^(r-1)# (for constant #C#) So, for all #s != 0#, the antiderivative of #x^s# is #x^(s+1)/(s+1)+C#. #1/x^2 = x^-2# so its antiderivative is #x^(-2+1)/(-2+1) = x^-1/-1 = -1/x# don't forget the #+C#. Answer link Related questions How do you evaluate the integral #intx^3+4x^2+5 dx#? How do you evaluate the integral #int(1+x)^2 dx#? How do you evaluate the integral #int8x+3 dx#? How do you evaluate the integral #intx^10-6x^5+2x^3 dx#? What is the integral of a constant? What is the antiderivative of the distance function? What is the integral of #|x|#? What is the integral of #3x#? What is the integral of #4x^3#? What is the integral of #sqrt(1-x^2)#? See all questions in Integrals of Polynomial functions Impact of this question 1324 views around the world You can reuse this answer Creative Commons License