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Find: ∫x2(x2+1)(3x2+4)dx

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प्रश्न

Find: `int x^2/((x^2 + 1)(3x^2 + 4))dx`

योग
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उत्तर

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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2021-2022 (March) Term 2 - Outside Delhi Set 1

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