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Question
Integrate the rational function:
`2/((1-x)(1+x^2))`
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Solution
`2/((1 - x)(1 + x^2)) = A/(1 - x) = (Bx + C)/(1 + x^2)`
2 = A(1 + x2) + (1 - x) Bx + C
Put x = 1
2 = 2A + 0
⇒ A = 1
Put x = 0
2 = A + C
⇒ C = 1
Comparing the coefficients of x2 on both sides,
0 = A - B
⇒ B = A = 1
`therefore 2/((1 - x)(1 + x^2)) = 1/(1 - x) + (x + 1)/(1 + x^2)`
`= 1/(1 - x) + x/(1 + x^2) + 1/(1 + x^2)`
On integrating
`int 2/((1 - x)(1 + x^2)) dx`
`= int 1/(1 - x) dx + 1/2 int (2x)/(1 + x^2) dx + 1/(1 + x^2) dx`
`= - log abs (1 - x) + 1/2 log abs (1 + x^2) + tan^-1 x + C`
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