Single Variable Calculus, Chapter 5, Review Exercises, Section Review Exercises, Problem 22

Find the intergral $\displaystyle \int \frac{x+2}{\sqrt{x^2+4x}} dx$, if it exists.
If we let $u = x^2 + 4x$, then $du = (2x+4) dx$, so $\displaystyle (x+2) dx = \frac{du}{2}$. Thus,

$\begin{equation} \begin{aligned} \int \frac{x+2}{\sqrt{x^2+4x}} dx &= \int \frac{1}{\sqrt{x^2+4x}} x+ 2 dx\\ \\ \int \frac{x+2}{\sqrt{x^2+4x}} dx &= \int \frac{1}{\sqrt{u}} \frac{du}{2}\\ \\ \int \frac{x+2}{\sqrt{x^2+4x}} dx &= \frac{1}{2} \int u^{\frac{-1}{2}} du\\ \\ \int \frac{x+2}{\sqrt{x^2+4x}} dx &= \frac{1}{2} \cdot \frac{u^{-\frac{1}{2}}+1}{-\frac{1}{2}+1} + C\\ \\ \int \frac{x+2}{\sqrt{x^2+4x}} dx &= \frac{1}{2} \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C\\ \\ \int \frac{x+2}{\sqrt{x^2+4x}} dx &= \frac{1}{\cancel{2}} \cdot \cancel{2}u^{\frac{1}{2}} + C\\ \\ \int \frac{x+2}{\sqrt{x^2+4x}} dx &= u^{\frac{1}{2}} + C\\ \\ \int \frac{x+2}{\sqrt{x^2+4x}} dx &= \left( x^2 + 4x \right)^{\frac{1}{2}} + C \qquad \text{or} \qquad \int \frac{x+2}{\sqrt{x^2+4x}} dx = \sqrt{x^2+4x} + C \end{aligned} \end{equation}$