Question

Find: `int (3x +5)/(x^2+3x-18)dx.`

Answer 1

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"`