∫(3x-2)/((x-1)(x2+1)) dx
To determine this integral, first decompose the function into partial fractions.
Set (3x-2)/((x-1)(x2+1) = A/(x - 1) + (Bx + C)/(x2+1) and solve for A, B and C.
You would end up with A = 1/2, B = - 1/2 , C = 5/2
Substituting,
∫(3x-2)/((x-1)(x2+1)) dx = ∫(1/2)[1/(x-1) - x/(x2+1) + 5/(x2+1)] dx
= (1/2)[ln|x-1| - (1/2)ln (x2 +1) + 5 Arctan x] + C
