Question

Integrate the function:

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

Answer 1

Let `I = 1/((x^2 + 1)(x^2 + 4)) = 1/((y + 1)(y + 4))`

Where x² = y

`= A/(y + 1) + B/(y + 4)`

`=>  A(y + 4) + B(y + 1)`       ...(1)

Putting y = -1 in equation (1),

∴ 1 = A(- 1 + 4)

`=> A = 1/3`

Putting y = -4 in equation (1),

∴ 1 = B(- 4 + 1)

`=> B = - 1/3`

`therefore 1/((x^2 + 1)(x^2 + 4)) = 1/(3 (y + 1)) - 1/(3 (y + 4))`

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

Now, `I = int [1/ (3(x^2 + 1)) - 1/ (3(x^2 + 4))]  dx`

`= (1/3  tan^-1 x) - (1/3 xx 1/2 tan^-1  (x/2)) + C`

`= 1/3 tan^-1 x - 1/6 tan^-1  (x/2) + C`