\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

NOTE REFERENCE TO QUESTION 1

{\bf Question}

Verify the following integral.

$\ds\int_{-\infty}^{+\infty} \ds\frac{x^2
\,dx}{(x^2+1)^2(x^2+2x+2)}=\ds\frac{7\pi}{50}$


\medskip

{\bf Answer}

Consider $\ds\oint_C \ds\frac{dz z^2}{(z^2+1)^2(z^2+2Z+2)}$ around
contour of Q1.

Integrand has poles at  $\begin{array} {rcl} z^2+1=0 & \Rightarrow
& z=\pm i\\ z^2+2z+2=0 & \Rightarrow & z=-1 \pm i \end{array}$

$z=+ i,\ z=-1+i$ are inside

$J=2\pi i \times$ [(residue at $+i$)+(residue at $-1+i$)]

\begin{eqnarray*} & & \rm{Residue\ at\ \un{double}\ pole} +i\\
& = & \lim_{z \to +i} \left[\ds\frac{1}{1!}\ds\frac{d}{dz}
\left\{\ds\frac{(z-i)^2z^2}{(z-i)^2(z+i)^2(z^2+2z+2)}\right\}\right]\\
& = & \lim_{z \to +i}
\left[\ds\frac{d}{dz}\left\{\ds\frac{z^2}{(z+i)^2(z^2+2z+1)}\right\}\right]\\
& = & \ds\frac{9i-12}{100} \end{eqnarray*}

\begin{eqnarray*} & & \rm{Residue\ at\ \un{simple}\ pole} -1+i\\
& = & \lim_{z \to -1+i}
\left[\ds\frac{(z+1-i)z^2}{(z+1)^2(z+1-i)(z-1+i)}\right]\\ & = &
\ds\frac{3-4i}{25} \end{eqnarray*}

Thus $J=2\pi i
\left(\ds\frac{9i-12}{100}+\ds\frac{3-4i}{25}\right)=\ds\frac{7\pi}{50}$

Now look at $R \to \infty$. Contribution around semicircle
vanishes to give

$$\ds\int_{-\infty}^{+\infty} \ds\frac{dx
x^2}{(x^2+1)^2(x^2+2x+2)}=\ds\frac{7\pi}{50}$$

as required.

\end{document}
