Answer
$$\int\frac{dx}{x\sqrt{x+4}}=\frac{1}{2}\ln\Big|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}\Big|+C$$
Work Step by Step
$$A=\int\frac{dx}{x\sqrt{x+4}}$$
Using Formula 29a, which states that
$$\int\frac{dx}{x\sqrt{ax+b}}=\frac{1}{\sqrt b}\ln\Big|\frac{\sqrt{ax+b}-\sqrt b}{\sqrt{ax+b}+\sqrt b}\Big|+C$$
with $a=1$ and $b=4$ here, we have
$$A=\frac{1}{\sqrt4}\ln\Big|\frac{\sqrt{x+4}-\sqrt 4}{\sqrt{x+4}+\sqrt 4}\Big|+C$$ $$A=\frac{1}{2}\ln\Big|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}\Big|+C$$
