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Question
Integrate the function `(x + 3)/(x^2 - 2x - 5)`
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Solution
Let `I = int (x + 3)/ (x^2 - 2x - 5) dx`
Put `x + 3 = A [d/dx (x^2 - 2x - 5)] + B`
= A (2x - 2) +B ....(i)
Comparing the coefficient of x in (i), we get
1 = 2A
⇒ `A = 1/2`
Comparing the constant terms in (i), we get
3 = B - 2A
⇒ 3 = B - 1
⇒ B = 4
`I = int (1/2 (2x - 2) + 4)/(x^2 - 2x - 5)`
`1/2 int (2x - 2)/ (x^2 - 2x - 5) dx + 4 int dx /(x^2 - 2x - 5)`
Let `I = 1/2 I_1 + 4I_2` ....(ii)
Where `I_1 = int (2x - 2)/ (x^2 - 2x -5) dx`
Put x2 - 2x - 5 = t
⇒ (2x - 2) dx = dt
∴ `I_1 = intdt/t = log |t| = log |x^2 - 2x - 5| + C_1` ....(iii)
and `I_2 = int dx/ (x^2 - 2x - 5)`
`= dx/ ((x - 1)^2 - 6)`
`= int dx/ ((x - 1)^2 - (sqrt6)^2) = 1/(2sqrt6) log |(x - 1 - sqrt6)/(x - 1 + sqrt6)| + C_2`
Hence from (ii), (iii) and (iv), we get
∴ `I = 1/2 log |(x^2 - 2x - 5)| + 2/ sqrt6 log |(x - 1 - sqrt6)/(x - 1 + sqrt6)| + C`
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