Question

How do you find the integral of `int 1/(4y-1) dy` from 0 to 1?

Answer 1

`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`