Use Of Partial Fraction? Really?

Calculus Level 4

∫ 0 ∞ x 2 ( 12 x 2 − 1 ) ( 4 x 2 + 1 ) 3 ( 1 + x 2 ) d x = π A \large \int_0^\infty \dfrac{x^2(12x^2 - 1)}{(4x^2 + 1)^3 (1+x^2)} \, dx = \dfrac \pi A

If the equation above holds true for some constant A A , find A A .


The answer is 54.

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Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Chew-Seong Cheong
Apr 18, 2016

As the title suggests, the problem can be solved by partial fraction. Let...

x 2 ( 12 x 2 − 1 ) ( 4 x 2 + 1 ) 3 ( 1 + x 2 ) = a ( 4 x 2 + 1 ) 3 + b ( 4 x 2 + 1 ) 2 + c 4 x 2 + 1 + d 1 + x 2 ⇒ x 2 ( 12 x 2 − 1 ) = a ( 1 + x 2 ) + b ( 4 x 2 + 1 ) ( 1 + x 2 ) + c ( 4 x 2 + 1 ) 2 ( 1 + x 2 ) + d ( 4 x 2 + 1 ) 3 \begin{aligned} \frac{x^2(12x^2-1)}{(4x^2+1)^3(1+x^2)} & = \frac{a}{(4x^2+1)^3} + \frac{b}{(4x^2+1)^2} + \frac{c}{4x^2+1} + \frac{d}{1+x^2} \\ \Rightarrow x^2(12x^2-1) & = a(1+x^2) + b(4x^2+1)(1+x^2) + c(4x^2+1)^2(1+x^2) + d(4x^2+1)^3 \end{aligned}

x 2 = − 1 4 : 3 4 a + b ( 0 ) + c ( 0 ) + d ( 0 ) = − 1 4 ( − 4 ) = 1 ⇒ a = 4 3 x 2 = − 1 : − 27 d = 13 ⇒ d = − 13 27 x = 0 : a + b + c + d = 0 ⇒ b + c = − 23 27 x = 1 : 2 a + 10 b + 50 c + 125 d = 11 ⇒ 10 b + 50 c = 1850 27 ⇒ b = − 25 9 ⇒ c = 52 27 \begin{aligned} x^2 = -\frac{1}{4}: \quad \frac{3}{4}a + b(0) + c(0) + d(0) & = - \frac{1}{4} (-4) = 1 \\ \Rightarrow a & = \frac{4}{3} \\ x^2 = -1: \quad -27d & = 13 \\ \Rightarrow d & = -\frac{13}{27} \\ x = 0: \quad a+b+c+d & = 0 \\ \Rightarrow b+c & = -\frac{23}{27} \\ x = 1: \quad 2a+10b+50c+125d & = 11 \\ \Rightarrow 10b+50c & = \frac{1850}{27} \\ \Rightarrow b & = - \frac{25}{9} \\ \Rightarrow c & = \frac{52}{27} \end{aligned}

Therefore, we have:

I = ∫ 0 ∞ x 2 ( 12 x 2 − 1 ) ( 4 x 2 + 1 ) 3 ( 1 + x 2 ) d x = ∫ 0 ∞ ( 4 3 ( 4 x 2 + 1 ) 3 − 25 9 ( 4 x 2 + 1 ) 2 + 52 27 ( 4 x 2 + 1 ) − 13 27 ( 1 + x 2 ) ) d x = ∫ 0 π 2 ( 2 3 sec ⁡ 4 θ − 25 18 sec ⁡ 2 θ + 26 27 − 13 27 ) d θ = ∫ 0 π 2 ( 2 cos ⁡ 4 θ 3 − 25 cos ⁡ 2 θ 18 + 13 27 ) d θ = 1 54 ∫ 0 π 2 ( 36 cos ⁡ 4 θ − 75 cos ⁡ 2 θ + 26 ) d θ = 1 54 ∫ 0 π 2 [ ( 36 cos ⁡ 4 θ − 36 cos ⁡ 2 θ + 9 ) − 39 cos ⁡ 2 θ + 17 ] d θ = 1 54 ∫ 0 π 2 [ 9 cos ⁡ 2 ( 2 θ ) − 39 2 cos ⁡ ( 2 θ ) + 1 ) + 17 ] d θ = 1 54 ∫ 0 π 2 [ 9 2 ( cos ⁡ ( 4 θ ) + 1 ) − 39 2 ( cos ⁡ ( 2 θ ) + 1 ) + 17 ] d θ = 1 54 ∫ 0 π 2 [ 9 2 cos ⁡ ( 4 θ ) − 39 2 cos ⁡ ( 2 θ ) + 2 ] d θ = 1 54 [ − 9 8 sin ⁡ ( 4 θ ) + 39 4 sin ⁡ ( 2 θ ) + 2 θ ] 0 π 2 = π 54 \begin{aligned} I & = \int_0^\infty \frac{x^2(12x^2-1)}{(4x^2+1)^3(1+x^2)} \, dx \\ & = \int_0^\infty \left(\frac{4}{3(4x^2+1)^3} - \frac{25}{9(4x^2+1)^2} + \frac{52}{27(4x^2+1)} - \frac{13}{27(1+x^2)} \right) \, dx \\ & = \int_0^\frac{\pi}{2} \left(\frac{2}{3\sec^4 \theta} - \frac{25}{18\sec^2 \theta} + \frac{26}{27} - \frac{13}{27} \right) \, d \theta \\ & = \int_0^\frac{\pi}{2} \left(\frac{2\cos^4 \theta}{3} - \frac{25\cos^2 \theta}{18} + \frac{13}{27} \right) \, d \theta \\ & = \frac{1}{54} \int_0^\frac{\pi}{2} \left(36\cos^4 \theta - 75\cos^2 \theta + 26 \right) \, d \theta \\ & = \frac{1}{54} \int_0^\frac{\pi}{2} \left[(36\cos^4 \theta - 36\cos^2 \theta + 9) - 39\cos^2 \theta +17 \right] \, d \theta \\ & = \frac{1}{54} \int_0^\frac{\pi}{2} \left[9\cos^2 (2\theta) - \frac{39}{2}\cos (2\theta)+1) + 17 \right] \, d \theta \\ & = \frac{1}{54} \int_0^\frac{\pi}{2} \left[\frac{9}{2}(\cos(4\theta) +1) - \frac{39}{2}(\cos (2\theta)+1) + 17 \right] \, d \theta \\ & = \frac{1}{54} \int_0^\frac{\pi}{2} \left[\frac{9}{2}\cos(4\theta) - \frac{39}{2}\cos (2\theta) + 2 \right] \, d \theta \\ & = \frac{1}{54} \left[ - \frac{9}{8}\sin(4\theta) + \frac{39}{4}\sin (2\theta) +2 \theta \right]_0^\frac{\pi}{2} \\ & = \frac{\pi}{54} \end{aligned}

⇒ A = 54 \Rightarrow A = \boxed{54}

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