Answer
$\ln 2$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{-1}^{\infty} \dfrac{d\theta}{ {\theta^2+5\theta+6}}=\lim\limits_{p \to \infty}\int_{-1}^{p} \dfrac{d\theta}{ {\theta^2+5\theta+6}}$ ....(1)
Then, we have
$\lim\limits_{p \to \infty}\int_{-1}^{p} \dfrac{d\theta}{ {\theta^2+5\theta+6}}=\lim\limits_{p \to \infty}\int_{-1}^{p}[\dfrac{1}{\theta+2}-\dfrac{1}{\theta+3}] $
or,
$\lim\limits_{p \to \infty}\int_{-1}^{p}[\dfrac{1}{\theta+2}-\dfrac{1}{\theta+3}]=\lim\limits_{p \to \infty}[\ln \dfrac{\theta+2}{\theta+3}]_{-1}^{p}$
Therefore, equation (1) becomes
$\lim\limits_{p \to \infty}[\ln \dfrac{\theta+2}{\theta+3}]_{-1}^{p}=\lim\limits_{p \to \infty}[\ln \dfrac{p+2}{p+3}-[\ln \dfrac{-1+2}{-1+3}]]=\ln 2$
