Chapter 5 - Review - Exercises - Page 423: 27

$\int \frac{x+2}{\sqrt{x^2+4x}}~dx = \sqrt{x^2+4x}+C$

$\int \frac{x+2}{\sqrt{x^2+4x}}~dx$ Let $u = x^2+4x$ $\frac{du}{dx} = 2x+4$ $dx = \frac{du}{2x+4}$ $\int \frac{x+2}{\sqrt{u}}~\frac{du}{2x+4}$ $=\int \frac{1}{2}~\frac{du}{\sqrt{u}}$ $= \sqrt{u}+C$ $= \sqrt{x^2+4x}+C$