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Question
Integrate the rational function:
`(x^3 + x + 1)/(x^2 -1)`
Solution
Let `I = int (x^3 + x + 1)/(x^2 - 1) dx`
Since `(x^3 + x + 1)/(x^2 - 1)` is an improper fraction, we convert it into a proper fraction by long division method.
`x^2 - 1) overline (x^3 + x + 1)(x`
x3 - x
- +
2x + 1
`(x^3 + x + 1)/(x^2 -1) = Q + R/D`
∴ `(x^3 + x + 1)/(x^2 - 1) = x + (2x + 1)/(x^2 - 1)` ....(i)
Now,
`(2x + 1)/(x^2 - 1) = (2x + 1)/ ((x + 1)(x - 1))`
`= A/(x + 1) + B/(x - 1)`
⇒ 2x + 1 = A (x - 1) + B (x + 1) ....(ii)
Putting x = -1 in (ii), we get
-2 + 1 = A (-1-1)
⇒ `A = (-1)/-2 = 1/2`
Putting x = 1 in (ii), we get
2 + 1 = B (1 + 1)
⇒ `B = 3/2`
∴ `(2x + 1)/(x^2 - 1) = 1/(2 (x + 1)) + 3/ (2 (x - 1))` ....(iii)
From (i) and (iii),
`(x^3 + x + 1)/(x^2 - 1) = x + 1/ (2(x + 1)) + 3/ (2 (x - 1))`
∴ `int(x^3 + x + 1)/(x^2 - 1) dx`
`= int x dx + 1/2 int dx/ (x + 1) + 3/2 int dx/ (x - 1)`
`= x^2/2 + 1/2 log |x + 1| + 3/2 log |x - 1| + C`
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