Question

Integrate the following w.r.t. x:

`(6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1)`

Answer 1

Let I = `int (6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1).dx`

`3x^2 - 2x - 1")"overline(6x^3 + 5x^2 -  7)("2x + 3`
                        `6x^3 - 4x^2 - 2x`
                         –     +        +           
                               `9x^2 + 2x - 7`
                               `9x^2 - 6x - 3`
                                –     +       +     
                                           8x  –  4

∴ I = `int [(2x + 3) + (8x - 4)/(3x^2 - 2x - 1)] dx`

= `int 2x + 3 + int (8x - 4)/((x - 1)(3x + 1)) dx`

Let `(8x - 4)/((x - 1)(3x + 1)) = "A"/(x - 1) + "B"/(3x + 1)`

∴ 8x – 4 = A(3x + 1) + B(x – 1)

Put x – 1 = 0, i.e. x = 1, we get

8 – 4 = A(4) + B(0)

∴ A = 1

Put 3x + 1 = 0, i.e. x = `-(1)/(3)`, we get

`8(-1/3) - 4 = "A"(0) + "B"(-4/3)`

∴ `(-8 - 12)/(3) = -(4"B")/(3)`

∴ B = 5

∴ `(8x - 4)/((x - 1)(3x + 1)) = (1)/(x - 1) + (5)/(3x + 1)`

∴ `I = 2 int x dx + 3 int 1 dx + int [(1)/(x - 1) + (5)/(3x + 1)] dx`

= `2 (x^2/2) + 3x + int (1)/(x - 1) dx + 5 int (1)/(3x + 1) dx`

= `x^2 + 3x + log |x - 1| + 5/3 log |3x + 1| + c`.