d/dx (x^2/x-1) = 2x(x - 1)^-1 - x^2(x -1)^-2 = [2x^2 - 2x - x^2] / [x-1]^2 = [x^2 - 2x] / [x - 1]^2 =
[x^2 -2x + 1 - 1] /[x -1]^2 = [x -1]^2/ [x -1]^2 - 1/[x - 1]^2 = 1 - 1/[x -1]^2 = 2 g(x)
So
g(x) = 1/2 - 1/ [2(x -1)^2]
And (1/3)g(x) = 1/6 - 1/[6(x -1)^2] ......... . so we have
3 3
∫ 1/6 dx - (1/6) ∫ 1/ (x - 1)^2 dx
2 2
The first integral evaluates to 1/6
For the second integral let x - 1 = u .....so...... du = dx
So we have
3
(1/6)∫ [1/ 1/u^2 ] du =
2
3
-(1/6)u^-1 ] = -(1/6)[ 1/ (3-1) - 1/(2-1)] = (1/6)[1 - 1/2] = 1/12
2
So 1/6 - 1/12 = 1/12
