Answer
$$\ln \left|\frac{6}{5}\right|$$
Work Step by Step
Given $$\int_{1}^{\infty} \frac{d x}{(x+2)(2 x+3)}$$
Since
\begin{aligned}
\frac{1}{(x+2)(2 x+3)} &=\frac{A}{(x+2)}+\frac{B}{(2 x+3)} \\
1 &=A(2 x+3)+B(x+2)
\end{aligned}
At $x=-2$, $B= 2$, and at $x= -3/2$ $A= -1$. Thus:
\begin{aligned}
\int_{1}^{\infty} \frac{d x}{(x+2)(2 x+3)}&=\lim_{R\to\infty}\int_{1}^{R} \frac{d x}{(x+2)(2 x+3)} \\
&=\lim_{R\to\infty}\int_{1}^{R}\left[\frac{2}{(2 x+3)}-\frac{1}{(x+2)}\right] d x \\
&=\lim_{R\to\infty}[\ln |2 x+3|-\ln |x+2|]_{1}^{R}\\
&=\lim_{R\to\infty}\ln \left|\frac{2 R+3}{R+2}\right|-\ln \left|\frac{5}{3}\right|\\
&=\ln |2|-\ln \left|\frac{5}{3}\right|\\
&=\ln \left|\frac{6}{5}\right|
\end{aligned}
