Question

Integrate the following w.r.t. x: `(2x^2 - 1)/(x^4 + 9x^2 + 20)`

Answer 1

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`.