Answer
$$\ln 2$$
Work Step by Step
\begin{align*}
\int_{-1}^{\infty} \frac{d \theta}{\theta^{2}+5 \theta+6}&=\int_{-1}^{\infty} \frac{d \theta}{ \theta+2}-\int_{-1}^{\infty} \frac{d \theta}{\theta+3}\\
&=\lim _{b \rightarrow \infty}\int_{-1}^{b} \frac{d \theta}{ \theta+2}-\lim _{b \rightarrow \infty}\int_{-1}^{b} \frac{d \theta}{\theta+3}\\
&=\lim _{b \rightarrow \infty}\left[\ln {|\theta+2|}-\ln {|\theta+3|} \right]\bigg|_{-1}^{b}\\
&=\lim _{b \rightarrow \infty}\left[\ln \left|\frac{\theta+2}{\theta+3}\right|\right]\bigg|_{-1}^{b}\\
&=\lim _{b \rightarrow \infty}\left[\ln \left|\frac{b+2}{b+3}\right|-\ln \left|\frac{-1+2}{-1+3}\right|\right]\\
&=0-\ln \left(\frac{1}{2}\right)=\ln 2
\end{align*}
