Answer
$\sqrt 3$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{0}^{1} \dfrac{\theta +1}{\sqrt {\theta^2+2\theta}}=\lim\limits_{p \to 0^{+}}\int_{p}^{1} \dfrac{\theta +1}{\sqrt {\theta^2+2\theta}}$ ....(1)
Plug $u(\theta)=\theta^2+2\theta \implies \dfrac{du}{2}=(\theta+1) d\theta$ and $u(p)=p^2+2b; u(1)=3$
Then, we have $\lim\limits_{p \to 0^{+}}(\dfrac{1}{2}) \int_{p^2+2p}^{3} \dfrac{du}{\sqrt u} d\theta=[\sqrt u]_{p^2+2p}^{3} $
Thus, equation (1) becomes
$\lim\limits_{p \to 0^{+}}[\sqrt 3-\sqrt{p^2+2p}]=\sqrt 3$
