Easy or difficult?

$\displaystyle \int _{ -\infty }^{ \infty }{ \frac { dx }{ x^{ 4 }+8x^{ 3 }+32x^{ 2 }+64x+64 } }$

$\displaystyle =\int _{ -\infty }^{ \infty }{ \frac { dx }{ x^{ 4 }+8x^{ 3 }+16x^{ 2 }+16x^{2}+64x+64 } }$

$\displaystyle =\int _{ -\infty }^{ \infty }{ \frac { dx }{ x^{ 2 }\left( x+4 \right)^{2} +16\left( x+2 \right)^{2} } }$

Let $\displaystyle u=x+2$

$\displaystyle =\int _{ -\infty }^{ \infty }{ \frac { du }{ \left( u-2 \right) ^{ 2 }\left( u+2 \right)^{2} +16\left( u \right)^{2} } }$

$\displaystyle =\int _{ -\infty }^{ \infty }{ \frac { du }{ \left( u^{2}-4 \right)^{2} +16u^{2} } }$

$\displaystyle =\int _{ -\infty }^{ \infty }{ \frac { du }{ u^{4}-8u^{2}+16 +16u^{2} } }$

$\displaystyle =\int _{ -\infty }^{ \infty }{ \frac { du }{ u^{4}+8u^{2}+16 } }$

$\displaystyle =\int _{ -\infty }^{ \infty }{ \frac { du }{\left( u^{2}+4 \right)^{2} } }$

Let $\displaystyle u=2\tan \theta$

$\displaystyle =\int _{ -\pi/2 }^{ \pi/2 }{ \frac { 2\sec^{2} \theta d\theta }{\left( 4 \tan^{2} \theta+4 \right)^{2} } }$

$\displaystyle =\int _{ -\pi/2 }^{ \pi/2 }{ \frac { 2\sec^{2} \theta d\theta }{16\sec^{4} \theta } }$

$\displaystyle =\int _{ -\pi/2 }^{ \pi/2 }{ \frac { 1 }{8} } \cos^{2} \theta d\theta$

$\displaystyle = \frac { 1 }{8} \int _{ -\pi/2 }^{ \pi/2 }\frac { 1+\cos2\theta }{ 2 } { d\theta }$

$\displaystyle =\frac { 1 }{ 8 } \frac { \theta +\sin \theta \cos \theta }{ 2 } | \begin{matrix} \pi /2 \\ -\pi /2 \end{matrix}$

$\displaystyle =\frac { 1 }{ 8 } \frac { \pi }{ 2 } = \boxed{\frac{\pi}{16}}$

So

$\displaystyle A=\boxed{16}$

$\begin{aligned} \int_{-\infty}^{\infty} \frac{dx}{x^4+8x^3+32x^2+64x+64} &= \int_{-\infty}^{\infty} \frac{1}{(x^2+4x+8)^2}\,dx \\&= \int_{-\infty}^{\infty} \frac{1}{((x+2)^2+4)^2} \quad\quad\quad\quad\quad\quad\quad\quad{\text{Apply integral substitution: } \int f(g(x)) \cdot g'(x) = \int f(u)\,du \space u = g(x)} \\&= \frac{1}{8} \int_{-\infty}^{\infty} \frac{1}{(v^2+1)^2} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad{u = 2v; du=2dv} \\&= \frac{1}{8} \int_{-\infty}^{\infty} \frac{\sec^2(w)}{(\tan^2(w)+1)^2}\,dw \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad{v=\tan(w); dv=\frac{1}{\cos^2(w)}dw}\\&= \frac{1}{8} \int_{-\infty}^{\infty} \frac{\sec^2(w)}{(\sec^2(w))^2}\,dw \\&= \frac{1}{8} \int_{-\infty}^{\infty} \frac{1+\cos(2w)}{2}\,dw \\&=\frac{1}{16} \int_{-\infty}^{\infty} 1+\cos(2w)\,dw \\&= \frac{1}{16} \int_{-\infty}^{\infty} \left( w + \frac{1}{2}\sin(2w)\right) \\&=\left[ \frac{1}{16}\left( \tan^{-1}\left( \frac{x+2}{2}\right)+\frac{1}{2}\sin\left(2\tan^{-1} \left( \frac{x+2}{2}\right) \right)\right)\right]_{-\infty}^{\infty} \\&=\frac{\pi}{32} - \left(-\frac{\pi}{32}\right) \space \square \end{aligned}$

$\LARGE \text{ADIOS!!!}$