# ∫dxx3-1 - Mathematics and Statistics

Sum

int ("d"x)/(x^3 - 1)

#### Solution

Let I = int ("d"x)/(x^3 - 1)

= int 1/((x - 1)(x^2 + x + 1))  "d"x

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)  .......(i)

Putting x = 1 in (i), we get

1 = A(12 + 1 + 1)

∴ 1 = 3A

∴  A = 1/3

Putting x = 0 in (i), we get

1 = A(0 + 0 + 1) + (0 + C)(0 – 1)

∴ 1 = A – C

∴ 1 = 1/3 - "C"

∴ C = - 2/3

Putting x = 2 in (i), we get

1 = A(22 + 2 + 1) + (2B + C)(2 – 1)

∴ 1 = 7A + 2B + C

∴ 1 = 7/3  + 2"B" - 2/3

∴ 1 = 5/3 + 2"B"

∴ (-2)/(3) = 2B

∴ B = -1/3

∴ I = int ((1/3)/(x - 1) + (-1/3x - 2/3)/(x^2 + x + 1))  "d"x

= 1/3 int(1/(x - 1) - (x + 2)/(x^2 + x + 1))  "d"x

= 1/3 int 1/(x - 1)  "d"x - 1/3 int (x + 2)/(x^2 + x + 1)  "d"x

= 1/3 int 1/(x - 1)  "d"x - 1/3*1/2 int (2x + 4)/(x^2 + x + 1)  "d"x

= 1/3 int 1/(x  1)  "d"x - 1/6 int ((2x + 1) + 3)/(x^2 + x + 1)*  "d"x

= 1/3 int 1/(x - 1)  "d"x - 1/6 int (2x + 1)/(x^2 + x + 1)  "d"x - 1/2 int  ("d"x)/(x^2 + x + 1)

= 1/3 log|x - 1| - 1/6 log|x^2 + x + 1| - 1/2 int ("d"x)/(x^2 + x + 1/4 - 1/4 + 1)     ......[∵  int ("f'"(x))/("f"(x))  "d"x = log|"f"(x)| + "c"]

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

= 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"

∴ I = 1/3  log|x - 1| - 1/6  log|x^2 + x + 1| - 1/sqrt(3) tan^-1 ((2x + 1)/sqrt(3)) + "c"

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Chapter 2.3: Indefinite Integration - Long Answers III

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