Integrate the function: $\frac{1}{\sqrt{9x^{2}+6x+5}}$

(A) To integrate $\int \frac{1}{\sqrt{9x^{2}+6x+5}} dx$,we first complete the square for the quadratic expression inside the square root:
$9x^{2}+6x+5 = (3x)^{2} + 2(3x)(1) + 1^{2} + 4 = (3x+1)^{2} + 2^{2}$.
Now,the integral becomes $\int \frac{1}{\sqrt{(3x+1)^{2} + 2^{2}}} dx$.
Let $t = 3x+1$,then $dt = 3dx$,which implies $dx = \frac{1}{3} dt$.
Substituting these into the integral:
$\frac{1}{3} \int \frac{1}{\sqrt{t^{2} + 2^{2}}} dt$.
Using the standard formula $\int \frac{1}{\sqrt{x^{2}+a^{2}}} dx = \log |x + \sqrt{x^{2}+a^{2}}| + C$:
$= \frac{1}{3} \log |t + \sqrt{t^{2} + 2^{2}}| + C$.
Substituting $t = 3x+1$ back:
$= \frac{1}{3} \log |(3x+1) + \sqrt{(3x+1)^{2} + 4}| + C$.
$= \frac{1}{3} \log |(3x+1) + \sqrt{9x^{2}+6x+5}| + C$,where $C$ is an arbitrary constant.