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Question
Integrate the rational function:
`x/((x^2+1)(x - 1))`
Solution
Let `x/((x^2 + 1)(x - 1)) = (Ax + B)/(x^2 + 1) + C/((x - 1))`
⇒ x = (A) (+ B)(x - 1) + C = `1/2`
Put x = 1
1 = 0 + 2C
⇒ C `= 1/2`
On comparing the coefficients of x2 or x
0 = A + C
⇒ A = `- 1/2`
and 1 = - A + B
⇒ B `= 1/2`
Hence, `int x/((x^2 + 1)(x - 1)) dx`
`= int (- 1/2 x + 1/2)/(x^2 + 1) dx + 1/2 int 1/(x - 1) dx`
`= -1/2 int (x - 1)/(x^2 + 1) dx + 1/2 log abs (x - 1) + C`
`= 1/4 int (2x)/(x^2 + 1) + 1/2 int 1/(x^2 + 1) dx + 1/2 log abs (x - 1) + C`
`= - 1/4 log abs (x^2 + 1) + 1/2 tan^-1 x + 1/2 log abs (x - 1) + C`
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