pleasee dear brainy

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