Area bounded by $y=\frac{1}{x^2-2x+2}$ and x –axis is

$$\begin{gathered} x^2-2x+2={\left(x-1\right)}^2+1\ge 1 \\ \Rightarrow \frac{1}{{\left(x-1\right)}^2+1}\le 1 \end{gathered}$$$X-$

$$\begin{gathered} and y=\frac{1}{x^2-2x+2} is symmetric about x=1 \\ and x\to \pm \infty \Rightarrow y\to 0 \end{gathered}$$

The area of the shaded region is the area of the reigon bounded by the curve $y=\frac{1}{x^2-2x+2}$ and $X-$ axis

This is equal to two times the area of the region bouned by the curve $y=\frac{1}{x^2-2x+2}$

 and the lines $x=0,x=\infty$ and $X-$ axis. 

Therefore, 

   $\begin{array}{rcl}A& =& {\left.2{\int }_1^{\infty } \frac{1}{(\mathrm{x}-1{)}^2+1}\mathrm{dx}=2{\tan }^{-1}(\mathrm{x}-1)\right]}_1^{\infty }\\ & =& 2\left(\frac{\pi }{2}-0\right)\\ & =& \mathrm{\pi }\end{array}$