Question

Integrate the rational function:

`((x^2 +1)(x^2 + 2))/((x^2 + 3)(x^2+ 4))`

Answer 1

`((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`