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Question
Integrate the rational function:
`((x^2 +1)(x^2 + 2))/((x^2 + 3)(x^2+ 4))`
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Solution
`((x^2 + 1)(x^2 + 2))/((x^2 + 3)(x^2 + 4))` Taking x2 = y
`((y + 1)(y + 2))/((y + 3)(y + 4)) = (y^2 + 3y + 2)/(y^2 + 7y + 12)`
`= 1 - (4y + 10)/(y^2 + 7y + 12)`
`= 1 - (4y + 10)/((y + 3)(y + 4))`
Let `(4y + 10)/((y + 3)(y + 4)) = A/((y + 3)) + B/(y + 4)`
4y + 10 = A (y + 4) + B (y + 3)
Putting y = -4 - 6 = 0 - B
⇒ B = 6
Putting y = -3, -2 = A + 0
⇒ A = -2
`therefore ((x^2 + 1)(x^2 + 2))/((x^2 + 3)(x^2 + 4)) = 1 - [(-2)/(y + 3) + 6/(y + 4)]`
`= 1 + 2/(y + 3) + 6/(y + 4)`
`int ((x^2 + 1)(x^2 + 2))/((x^2 + 3)(x^2 + 4))` dx
`= int dx + 2 int 1/(x^2 sqrt(3^2)) + 6 int 1/(x^2 + 4)` dx
`= x + 2/sqrt 3 tan^-1 x/sqrt3 - 6/2 tan^-1 (x/2) + C`
`= x + 2/sqrt 3 tan^-1 x/sqrt3 - 3 tan^-1 x/2 + C`
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