whats the integral of (x-3)/(2x-3) ?? What technique do u use?

Answer 1

u = 2x - 3
du = 2

x - 3 = 1/2(u) - 3/2

so we want to find the integral of
[1/2(u) - 3/2]/u * 1/2 du
= 1/2(1/2)(u - 3)/u du
= 1/4integral((u-3)/u du)
= 1/4 integral(u/u - 3/u) =
= 1/4 integral(1 - 3/u) =
= 1/4 int(1) - 1/4integral(3/u)
= 1/4 u - 3/4 ln (abs(u))
= 1/4 (2x - 3) - 3/4 ln(abs(2x -3)) + c

Answer 2

(x-3)/(2x-3) = -.5 -1.5/2x-3
So integral = -.5x -3ln 2x-3 +C