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प्रश्न
Integrate the following w.r.t. x: `(2x^2 - 1)/(x^4 + 9x^2 + 20)`
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उत्तर
Let I = `int (2x^2 - 1)/(x^4 + 9x^2 + 20).dx`
Consider, `(2x^2 - 1)/(x^4 + 9x^2 + 20)`
For finding partial fractions only, put x2 = t.
∴ `(2x^2 - 1)/(x^4 + 9x^2 + 20) = t/((t + 1)(t - 2)(t + 3)`
= `"A"/(t + 1) + "B"/(t - 2) + "C"/(t + 3)` ...(Say)
∴ t = A(t – 2)(t + 3) + B(t + 1)(t + 3) + C(t + 1)(t –2)
Put t + 1 = 0, i.e. t = – 1, we get
–1 = A(– 3)(2) + B(0)(2) + C(0)(– 3)
∴ – 1 = – 6A
∴ A = `(1)/(6)`
Put t – 2 = 0, i.e. t = 2, we get
2 = A(0)(5) + B(3)(5) + C(3)(0)
∴ 2 = 15B
∴ B = `(2)/(15)`
Put t + 3 = 0, i.e. t = – 3, we get
– 3 = A(– 5)(0) + B(– 2)(0) + C(– 2)(– 5)
–3 = 10C
∴ C = `-(3)/(10)`
∴ `t/((t + 1)(t - 2)(t + 3)) = ((1/6))/(t + 1) + ((2/15))/(x^2 - 2) + (((-3)/10))/(x^2 + 3)`
∴ `x^2/((x^2 + 1)(x^2 - 2)(x^2 + 3)) = ((1/6))/(x^2 + 1) + ((2/15))/(x^2 - 2) + (((-3)/10))/(x^2 + 3)`
∴ I = `int [((1/6))/(x^2 + 1) + ((2/15))/(x^2 - 2) + (((-3)/10))/(x^2 + 3)].dx`
= `(1)/(6) int (1)/(1 + x^2).dx + (2)/(15) int (1)/(x^2 - (sqrt(2))^2).dx - (3)/(10) int (1)/(x^2 + (sqrt(3))^2).dx`
= `(1)/(6) tan^-1 x + (2)/(15) xx (1)/(2sqrt(2))log|(x - sqrt(2))/(x + sqrt(2))| - (3)/(10) xx (1)/sqrt(3)tan^-1(x/sqrt(3)) + c`
= `(11)/sqrt(5)tan^-1 (x/sqrt(5)) - (9)/(2)tan^-1(x/2) + c`.
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