Improper Integral: Infinite limits

Firstly, $\begin{aligned}\frac{4x-30}{(x^2+5)(3x+2)}&\equiv\frac{Ax+B}{x^2+5}+\frac{C}{3x+2}\\&\equiv \frac{(Ax+B)(3x+2)+C(x^2+5)}{(x^2+5)(3x+2)}\\&\equiv \frac{(3A+C)x^2+(2A+3B)x+2B+5C}{(x^2+5)(3x+2)}\end{aligned}$

Comparing coefficients, we get $A=2,B=0$ and $C=-6$ .

$\begin{aligned}\int_0^\infty\frac{4x-30}{(x^2+5)(3x+2)}dx&=\int_0^\infty\frac{2x}{x^2+5}-\frac{6}{3x+2}dx\\&=\lim_{n\to\infty}\int_0^n\frac{2x}{x^2+5}-\frac{6}{3x+2}dx\\&=\lim_{n\to\infty}\left [ \ln|x^2+5|-2\ln|3x+2| \right ]_0^n\\&=\lim_{n\to\infty}\left [\ln\frac{x^2+5}{(3x+2)^2}\right ]_0^n\\&=\lim_{n\to\infty}\left[\ln\left (\frac{n^2+5}{9n^2+12n+4}\right )-\ln\left (\frac{5}{4}\right )\right ]\\&=\ln\frac{1}{9}-\ln\frac{5}{4}\\&=\ln\frac{4}{45}=\color{#20A900}\boxed{-2.42\ \textnormal{(3.s.f)}}\end{aligned}$