Answer
$$\sqrt{3}$$
Work Step by Step
Given
$$\int_{0}^{1} \frac{\theta+1}{\sqrt{\theta^{2}+2 \theta}} d \theta $$
Let
$u=\theta^{2}+2 \theta \ \to d u=2(\theta+1) d \theta $ and at $\theta = 0\to u=0$, $\theta = 1\to u=3$, then
\begin{align*}
\int_{0}^{1} \frac{\theta+1}{\sqrt{\theta^{2}+2 \theta}} d \theta &=\int_{0}^{3} \frac{d u}{2 \sqrt{u}}\\
&=\lim _{b \rightarrow 0^{+}} \int_{b}^{3} \frac{d u}{2 \sqrt{u}}\\
&=\lim _{b \rightarrow 0^{+}}[\sqrt{u}]_{b}^{3}\\
&=\lim _{b \rightarrow 0^{+}}(\sqrt{3}-\sqrt{b})\\
&=\sqrt{3}-0=\sqrt{3}
\end{align*}
