Question #07f78 Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Eddie Aug 12, 2016 Does not converge Explanation: Does it converge? We'll find out #int_3^oo1/(2x-1)dx# #= lim_{alpha to oo} int_3^alpha \ 1/(2x-1)dx# #= lim_{alpha to oo} int_3^alpha \ 1/2 d/dx ln(2x-1)dx# #= 1/2 lim_{alpha to oo} [ ln(2x-1) ]_3^alpha# #= 1/2 lim_{alpha to oo} ln((2alpha-1)/3)# #beta = (2alpha-1)/3# #= 1/2 lim_{beta to oo} ln beta# #= oo# 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 1055 views around the world You can reuse this answer Creative Commons License