Simplify (3x-2)/(x-1) as a sum of two terms. Hence find ∫(3x-2)/(x-1) dx

Answer 1

integral of

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

=integral of (3u+1)du/u

= integral of du/u + 3du

= lnu +3u + C

= ln|x-1| +3(x-1) + C

= ln|x-1| + 3x + C (combine the -3 and C into one constant term)

or

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

= 3x/(x-1) + -2/(x-1) let u = x-1, du=dx

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

= 3u/u +3/u -2/u

=3 +1/u

integral of 3du + du/u

= 3u + ln|u| + C

=3(x-1) + ln|x-1| + C

=3x -3 + ln|x-1| + C

= 3x + ln|x-1| + C