An Inverse polynomial integral

By Residue Theorem: As suggested by @Syed Shahabudeen

$\begin{aligned} I & = \int_{-\infty}^\infty \frac {dx}{x^6+3x^4+3x^2+1} = \int_{-\infty}^\infty \frac {dx}{(x^2+1)} \end{aligned}$

Using a big semicircular contour $C$ in the upper half complex plane, it will have length $\pi R$ , but the integrand of the order $1/R^3$ , so this contribution from the big semicircle goes to zero on taking $R \to \infty$ . Then we have

$\begin{aligned} I & = \int_C \frac 1{(z^2+1)^3} dz \\ & = \int_C \frac {dz}{(z+i)^3(z-i)^3} & \small \blue{\text{Only pole }z=i \text{ of }z=\pm i \text{ is enclosed by the contour.}} \\ & = \frac {2\pi i}{2!} \left[\frac {d^2}{dz^2} \left(\frac 1{(z+i)^3} \right) \right]_{z=i} & \small \blue{\text{By Cauchy integral formula}} \\ & = \frac {3\pi}8 \end{aligned}$

Therefore $A+B+C = 3+8+1 = \boxed{12}$ .

References:

• Contour integration
• Residue theorem
• Cauchy integral formula

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By Beta and Gamma Functions: Similar solution with @Syed Shahabudeen's

$\begin{aligned} I & = \int_{=\infty}^\infty \frac 1{x^6+3x^4+3x^2+1} dx & \small \blue{\text{Since the integrand is even}} \\ & = 2 \int_0^\infty \frac {dx}{(x^2+1)^3} & \small \blue{\text{Let }x = \tan \theta \implies dx = \sec^2 \theta \ d\theta} \\ & = 2 \int_0^\frac \pi 2 \frac {\sec^2 \theta}{\sec^4 \theta} d\theta \\ & = 2 \int_0^\frac \pi 2 \sin^0 \theta \cos^4 \theta d\theta & \small \blue{\text{Beta function }B(m,n) = 2\int_0^\frac \pi 2 \sin^{2m-1} t \cos^{2n-1} t\ dt} \\ & = B \left(\frac 12, \frac 52\right) & \small \blue{\text{and }B(m,n) = \frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)}} \\ & = \frac {\Gamma\left(\frac 12\right)\Gamma\left(\frac 52\right)}{\Gamma\left(3\right)} & \small \blue{\text{where }\Gamma (\cdot) \text{ denotes the gamma function.}} \\ & = \frac {\Gamma\left(\frac 12\right)\cdot \frac 32\cdot \frac 12 \Gamma\left(\frac 12\right)}{2!} & \small \blue{\text{From }\Gamma (1+s) = s\Gamma(s), \ \Gamma\left(\frac 12\right) = \sqrt \pi} \\ & = \frac {3\pi}8 & \small \blue{\text{and }\Gamma (n) = (n-1)!} \end{aligned}$

Therefore $A+B+C = 3+8+1 = \boxed{12}$ .

References:

• Beta function
• Gamma function

Since the 6 degree polynomial is just an expansion of $\left( 1+{ x }^{ 2 } \right) ^{ 3 }$ . The following integral can be written as $\int _{ -\infty }^{ \infty }{ \frac { 1 }{ { \left( 1+{ x }^{ 2 } \right) }^{ 3 } } } dx$ . Since the integrand is an even function. It can be written as $2\int _{ 0 }^{ \infty }{ \frac { 1 }{ { \left( 1+{ x }^{ 2 } \right) }^{ 3 } } } dx$ on substituting $x = \tan \theta$ we get $2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { \sec }^{ 2 }\theta }{ { \left( 1+{ \tan }^{ 2 }\theta \right) }^{ 3 } } } d\theta$ on simplifying we get $2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \cos }^{ 4 }\theta } d\theta$ By applying wallis formula for sine cosine special integrals. we'll get $2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \cos }^{ 4 }\theta } d\theta \quad =\quad 2\left( \frac { 1 }{ 2 } \left( \frac { \Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 5 }{ 2 } \right) }{ \Gamma \left( 3 \right) } \right) \right) =\frac { 3\pi }{ 8 }$ . Therefore we get $\boxed{A+B+C=12}$