How do you find the integral of #int lnx dx # from 0 to 4? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Dharma R. Sep 16, 2015 #8ln2-4# Explanation: #int_0^4# #lnxdx# by applying byparts #lnx##int_0^4##dx#-#int_0^4##1/x##xdx#(apply lmints for lnx also) =(#ln4*4-ln0*0)#-(#4-0#) = #8ln2-4# 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 1478 views around the world You can reuse this answer Creative Commons License