Question

Integrate the following w.r.t. x : `(1)/(x^3 - 1)`

Answer 1

Let I = `int (1)/(x^3 - 1)*dx`

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

Let `(1)/((x - 1)(x^2 + x + 1))  = "A"/(x - 1) + ("B"x + "C")/(x^2 + x + 1)`

∴ 1 = A(x² + x + 1) + (Bx + C)(x - 1)
Put x – 1  = 0 i.e x = 1, we get
1 = A(3) + (B + C)(0)

∴ A = `(1)/(3)`
Put x = 0, we get
1 = A(1) + C(– 1)
∴ C = A – 1 = `-(2)/(3)`
Comparing the coefficients of x² on both the sides, we get
0 = A + B
∴  B = – A = `-(1)/(3)`

∴ `(1)/((x - 1)(x^2  + x + 1)) = ((1/3))/(x - 1) + ((-1/3x - 2/3))/(x^2 + x + 1)`

= `(1)/(3)[1/(x - 1) - (x + 2)/(x^2 + x + 1)]`

Let x + 2 = `"p"[d/dx(x^2 + x + 1)] + "q"`
Comapring coefficient of x and the constant term on both the sides, we get

2p = 1 i.e. p = `(1)/(2) and p + q` = 2

∴ q = 2 – p = `2 - (1)/(2) = (3)/(2)`

∴ x + 2 =`(1)/(2)(2x + 1) + (3)/(2)`

∴ `1/((x + 1)(x^2 + x + 1)) = (1)/(3)[1/(x - 1) - ((1)/(2)(2x + 1) + 3/2)/((x^2 + x + 1))]`

= `(1)/(3)[1/(x - 1) - (1)/(2)((2x + 1)/(x^2 + x + 1)) - ((3/2))/(x^2 + x + 1)]`

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

= `(1)/(3) int 1/(x - 1)*dx - (1)/(6) int (2x + 1)/(x^2 + x + 1)*dx - (1)/(2) int (1)/(x^2 + x + 1/4 + 3/4)*dx`

= `(1)/(3)log|x - 1| - (1)/(6) int (d/dx(x^2 + x + 1))/(x^2 + x + 1)*dx - (1)/(2) int (1)/((x + 1/2)^2 + (sqrt(3)/2)^2)*dx`

= `(1)/(3)log|x - 1| - (1)/(6)log|x^2 + x + 1| - (1)/(2)(1)/((sqrt(3)/2))tan^-1[((x + 1/2))/((sqrt(3)/2))] + c`

= `(1)/(3)log|x - 1| - (1)/(6)log|x^2 + x + 1| - (1)/sqrt(3)tan^-1((2x + 1)/sqrt(3)) + c`.