There are shortcut methods, but this is the way I was taught to do partial fractions:
Assume that the given expression has the following decomposition:
[MATH]\frac{1}{(x+2)(2x+1)}=\frac{A}{x+2}+\frac{B}{2x+1}[/MATH]
Multiply by \((x+2)(2x+1)\) to get:
[MATH]1=A(2x+1)+B(x+2)[/MATH]
Distribute and arrange as:
[MATH]0x+1=(2A+B)x+(A+2B)[/MATH]
Equating corresponding coefficients, we obtain the system:
[MATH]2A+B=0[/MATH]
[MATH]A+2B=1[/MATH]
Solving this system we obtain:
[MATH](A,B)=\left(-\frac{1}{3},\frac{2}{3}\right)[/MATH]
And so we know:
[MATH]\frac{1}{(x+2)(2x+1)}=\frac{-\dfrac{1}{3}}{x+2}+\frac{\dfrac{2}{3}}{2x+1}=\frac{1}{3}\left(\frac{2}{2x+1}-\frac{1}{x+2}\right)[/MATH]
Back to the integral, suppose we write it as:
[MATH]I=\frac{1}{3}\lim_{t\to\infty}\left(\ln\left(\frac{2t+1}{t+2}\right)+\ln(2)\right)[/MATH]
[MATH]I=\frac{1}{3}\lim_{t\to\infty}\left(\ln\left(\frac{2(2t+1)}{t+2}\right)\right)[/MATH]
[MATH]I=\frac{1}{3}\lim_{t\to\infty}\left(\ln\left(\frac{2\left(2+\dfrac{1}{t}\right)}{1+\dfrac{2}{t}}\right)\right)[/MATH]
Can you proceed?