residue theorem, integral of inverse quadratic not working $$\int_{-\infty}^\infty\frac{dx}{3x^2+0.4x+10}$$ roots of $3x^2+0.4x+10$ $= z_1,z_2=-\frac{1}{15}\pm 1.825i$ Using $\mathrm{res} = \frac{1}{z_1-z_2}=-0.274i$ ans: $2\pi i\cdot -0.274i=1.72$ the correct answer is $0.574$. What am I doing wrong?

Consider the upper demi-circle : $C_R=\\{z \in\mathbb{C} |\Im(z)\geq 0, |z|=R\\} $ and the segment $I_R=[-R,R]$. For $R$ big enough we have :

$$\int_{C_R\cup I_R}\frac{dz}{3z^2+0.4z+10}=\frac{1}{3}\int_{C_R\cup I_R}\frac{dz}{(z-z_1)(z-z_2)}=\frac{2\pi i}{3}Res(\frac{1}{(z-z_1)(z-z_2)},z_1)=\frac{2\pi i}{3}\frac{1}{z_1-z_2}=\frac{2\pi i}{3}\frac{1}{2i1.825}=\frac{\pi }{1.825\times3}\approx0.574$$

$$\int_{C_R\cup I_R}\frac{dz}{3z^2+0.4z+10}=\int_{C_R}\frac{dz}{3z^2+0.4z+10}+\int_{I_R}\frac{dz}{3z^2+0.4z+10}$$

$$|\int_{C_R}\frac{dz}{3z^2+0.4z+10}|\le\int_{C_R}|\frac{dz}{3z^2+0.4z+10}|\le\frac{\pi R}{3R^2+0.4R+10}\xrightarrow[R\rightarrow\infty]{\text{}}0$$

$$\int_{I_R}\frac{dz}{3z^2+0.4z+10} \xrightarrow[R\rightarrow\infty]{\text{}}\int_{-\infty}^\infty\frac{dx}{3x^2+0.4x+10}$$

Finally : $\int_{-\infty}^\infty\frac{dx}{3x^2+0.4x+10}\approx0.574$.