How do you evaluate #int (-1/(3x)) dx# for [1/11, 1/5]? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Trevor Ryan. Oct 19, 2015 #0,26282# Explanation: #int_(1/11)^(1/5)-1/(3x)dx=-1/3int_(1/11)^(1/5)1/xdx# #=1/3[ln|x|]_(1/11)^(1/5)# #=1/3[ln(1/5)-ln(1/11)]# #=0,26282# 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 1386 views around the world You can reuse this answer Creative Commons License