The area is given by
[tex]\displaystyle\int_0^\infty\frac{\mathrm dx}{x^2+6x+10}[/tex]
Complete the square in the denominator:
[tex]x^2+6x+10=x^2+6x+9+1=(x+3)^2+1[/tex]
then substitute [tex]x+3=\tan y[/tex], so that [tex]\mathrm dx=\sec^2y\,\mathrm dy[/tex]. Then as [tex]x\to0^+[/tex], we have [tex]y\to\arctan3[/tex]; as [tex]x\to\infty[/tex], we have [tex]y\to\dfrac\pi2[/tex]. So the integral becomes
[tex]\displaystyle\int_{\arctan3}^{\pi/2}\frac{\sec^2y}{\tan^2y+1}\,\mathrm dy=\int_{\arctan3}^{\pi/2}\mathrm dy=\dfrac\pi2-\arctan3[/tex]
