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Question
Integrate the rational function:
`(3x + 5)/(x^3 - x^2 - x + 1)`
Solution
Let `(3x + 5)/(x^3 - x^2 - x + 1)`
`= (3x + 5)/(x^2(x - 1) - 1(x - 1))`
`= (3x + 5)/((x^2 - 1)(x - 1))`
`= (3x + 5)/((x + 1)(x - 1)^2)`
`(3x + 5)/((x + 1)(x - 1)^2) = A/(x + 1) = B/(x - 1) + C/((x - 1)^2)`
3x + 5 = A(x - 1)2 + B(x2 - 1) + C(x + 1) ... (i)
Put x = 1
8 = 0 + 0 + 2C
⇒ C = 4
Put x = -1
2 = A(-2)2 + 0 = 0
⇒ A = `-1/2`
On comparing the coefficients of x2
0 = A + B
⇒ A = -A `= 1/2`
Hence, `(3x + 5)/(x^3 - x^2 - x + 1)`
`= -1/(2(x + 1)) + 1/(2(x - 1)) + 4/((x - 1)^2)`
On integrating,
`int (3x + 5)/(x^3 - x^2 - x + 1)`
`= -1/2 int 1/(x + 1) dx + 1/2 int 1/(x - 1) dx + 4 int 1/((x - 1)^2) dx`
`= -1/2 log abs (x + 1) + 1/2 log (x - 1) + 4 (1/((x - 1))) + C`
`= 1/2 log abs ((x + 1)/(x - 1)) - 4/((x - 1)) + C`
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