Answer
$$\int \frac{dx}{(x+3)\sqrt{(x+3)^2-25}}=\frac{1}{5}\sec^{-1}\Big(\Big|\frac{x+3}{5}\Big|\Big)+C$$
Work Step by Step
$$A=\int \frac{dx}{(x+3)\sqrt{(x+3)^2-25}}$$
We set $a=x+3$, which means $$da=dx$$
Therefore, $$A=\int\frac{da}{a\sqrt{a^2-25}}=\int\frac{da}{a\sqrt{a^2-5^2}}$$
We have $$\int\frac{dx}{x\sqrt{x^2-u^2}}=\frac{1}{u}\sec^{-1}\Big|\frac{x}{u}\Big|+C$$
So, $$A=\frac{1}{5}\sec^{-1}\Big|\frac{a}{5}\Big|+C$$ $$A=\frac{1}{5}\sec^{-1}\Big(\Big|\frac{x+3}{5}\Big|\Big)+C$$
