What is the antiderivative of #lnx/x^2#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Bdub Apr 5, 2016 #int(lnx)/x^2 =-(lnx+1)/x# Explanation: #u=ln x, dv=1/x^2# #du=1/x, v=-1/x# #intudv=uv-intvdu# #=-lnx/x-int-1/x^2 dx# #=-lnx/x+int1/x^2 dx# #=-lnx/x-1/x# #=-(lnx+1)/x# 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 1187 views around the world You can reuse this answer Creative Commons License