How do you find the integral of #int 1/(4y-1) dy# from 0 to 1? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Sasha P. Sep 26, 2015 #ln3/4# Explanation: #4y-1=t => 4dy=dt => dy=dt/4# #t_1=4*0-1=-1# #t_2=4*1-1=3# #I=int_0^1 1/(4y-1)dy = int_-1^3 1/t dt/4 = 1/4 int_-1^3 dt/t = 1/4 ln|t| |_-1^3# #I=1/4(ln|3|-ln|-1|)=1/4(ln3-ln1)=1/4(ln3-0)# #I=ln3/4# 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 3691 views around the world You can reuse this answer Creative Commons License