**1 Answer**

written 3.8 years ago by | modified 3.6 years ago by |

Consider the contour C consisting of large semi circle and diameter on the real axis with centre at the origin lying in the upper half of the plane.

Consider $ \frac{(z^2+z+2)}{(z^4+10z^2+9)}$ ,

poles are given by putting

$z^4+10z^2+9=0$ i.e. $(z^2+1^2)(z^2+9) =0$ i.e. $z = +i, - i ,+3i , -3i.$

Out of these poles $z=i,z=3i$ lie in upper half of the plane.

Residue at $z= i$

$R_1 = lim_{z→i} \ (z-i) \frac{(z^2+z+2)}{(z-i)(z+i) (z^2+9)} = \frac{(1+i)}{16i} $

Similarly,

Residue at $z= 3i $

$R_2 = lim_{z→3i} \ (z-3i) \frac{(z^2+z+2)}{(z-3i)(z+3i) (z^2+9)} = \frac{(7-3i)}{48i}$

$∫_{-∞}^∞ \frac{(x^2+x+2)}{(x^4+10x^2+9)} \ dx = 2πi[\frac{(1+i)}{16i}+\frac{(7-3i)}{48i}] = \frac{5π}{12}$