# Integrate the following w.r.t. x : 1x3-1 - Mathematics and Statistics

Sum

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

#### Solution

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.

Is there an error in this question or solution?
Chapter 3: Indefinite Integration - Exercise 3.4 [Page 145]

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