Answer
$$\frac{1}{2}\ln \left| {\frac{{\sqrt {x + 4} - 2}}{{\sqrt {x + 4} + 2}}} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\sqrt {x + 4} }}} \cr
& {\text{integrate by using the table of integrals from book}} \cr
& {\text{by the formula 29}}\left( a \right):\,\,\,\,\,\,\,\int {\frac{{dx}}{{x\sqrt {ax + b} }}dx = \frac{1}{{\sqrt b }}\ln \left| {\frac{{\sqrt {ax + b} - \sqrt b }}{{\sqrt {ax + b} + \sqrt b }}} \right|} + C \cr
& {\text{setting }}a = 1{\text{ and }}b = 4{\text{ then}} \cr
& \int {\frac{{dx}}{{x\sqrt {x + 4} }}} = \frac{1}{{\sqrt 4 }}\ln \left| {\frac{{\sqrt {x + 4} - \sqrt 4 }}{{\sqrt {x + 4} + \sqrt 4 }}} \right| + C \cr
& {\text{simplifying}} \cr
& \int {\frac{{dx}}{{x\sqrt {x + 4} }}} = \frac{1}{2}\ln \left| {\frac{{\sqrt {x + 4} - 2}}{{\sqrt {x + 4} + 2}}} \right| + C \cr} $$
