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Question
Find: `int x^2/((x^2 + 1)(3x^2 + 4))dx`
Solution
Let I = `int x^2/((x^2 + 1)(3x^2 + 4))dx`
Put t = x2
`t/((t + 1)(3t + 4)) = A/(t + 1) + B/(3t + 4)`
t = A(3t + 4) + B(t + 1)
t = (3A + B)t + (4A + B)
On comparing both sides, we get
3A + B = 1 and 4A + B = 0
∴ I = `int (-1)/(x^2 + 1)dx + int (-4)/(3x^2 + 4)dx`
= `-int 1/(x^2 + 1)dx - 4int 1/(3x^2 + 4)dx`
= `-int 1/(x^2 + 1^2)dx - 4int 1/((sqrt(3)x)^2 + 2^2)dx`
= `(-1)/1 tan^-1 (x/2) - 4/(2sqrt(3)) tan^-1 ((sqrt(3)x)/2) + C`
= `-tan^-1x - 2/sqrt(3) tan^-1 ((sqrt(3)x)/2) + C`
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