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Question
Integrate the function:
`1/((x^2 + 1)(x^2 + 4))`
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Solution
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`
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