Answer
$$\sqrt{x^2+4x}+c$$
Work Step by Step
Given
$$\int \frac{x+2}{\sqrt{x^2+4x}}dx$$
Let $u=x^2+4x\ \Rightarrow \ du= 2(x+2)dx $ , then
\begin{align*}
\int \frac{x+2}{\sqrt{x^2+4x}}dx&=\frac{1}{2}\int \frac{du}{\sqrt{u}}\\
&=\sqrt{u}+c\\
&=\sqrt{x^2+4x}+c
\end{align*}
