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Question
Find: `int (3x +5)/(x^2+3x-18)dx.`
Sum
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Solution
Let `I = int ((3x+5))/(x^2 +3x -18)dx`
`I = int ((3x+5)dx)/((x+6) (x-3))`
let `(3x+5)/((x+6) (x-3)) = "A"/(x+6) + "B"/(x-3)`
so 3x + 5 = A (x -3) + B (x +6)
On comparing,
A + B = 3 ...(i)
-3A + 6B = 5 ...(ii)
-3A + 6(3 - A) = 5
-3A + 18 - 6A = 5
`"A" = -13/-9 = 13/9 and "B" = 3 - "A" = 3 - 13/9 = 14/9`
So, `int ((3x+5)dx)/((x+6)(x-3)) = int (13dx)/(9(x+6)) + int(14dx)/(9(x-3))`
= `13/9 "In" (x+6)+14/9"In"(x-3) + "C"`
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