A New Year Integral in Disguise

$\begin{aligned} I & = \int_0^\infty \frac {dx}{(x+1)(x^4+4x^3+6x^2+4x+1)^{505}} \\ & = \int_0^\infty \frac {dx}{(x+1)\left((x+1)^4\right)^{505}} \\ & = \int_0^\infty \frac {dx}{(x+1)^{2021}} & \small \blue{\text{Let }u = x+1 \implies du = dx} \\ & = \int_1^\infty \frac 1{u^{2021}}\ du \\ & = - \frac 1{2020 u^{2020}} \bigg|_1^\infty \\ & = \frac 1{2020} \end{aligned}$

Therefore $A=\boxed{2020}$ .

$\Large \red{\text{Happy New Year}}$

The 4th-degree polynomial is just an expansion of ${ \left( 1+{ x } \right) }^{ 4 }$ . So the following integral can be written as $\int _{ 0 }^{ \infty }{ \frac { 1 }{ \left( x+1 \right) { \left( { x }+1 \right) }^{ 4\times 505 } } dx } =\int _{ 0 }^{ \infty }{ \frac { 1 }{ { \left( { x }+1 \right) }^{ 2021 } } dx }$ On substituting $x={ \tan }^{ 2 }\theta$ ( $x+1=t$ is also a good substitution) we get $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 2\tan\theta { \sec }^{ 2 }\theta }{ { \left( 1+{ \tan }^{ 2 }\theta \right) }^{ 2021 } } d\theta }$ On simplifying and by applying wallis formula we get $2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin\theta { \cos }^{ 4039 }\theta d\theta } =2\left( \frac { 1 }{ 2 } \left( \frac { \Gamma \left( 2020 \right) \Gamma \left( 1 \right) }{ \Gamma \left( 2021 \right) } \right) \right) =\frac { 1 }{ 2020 }$ therefore $\boxed{A=2020}$