Find values of constants a and b such that the given improper integral converges Find values of constants a and b such that:$$\int_{3}^\infty \left(\frac{ax+2}{x^2+3x}-\frac{b}{3x-2}\right) dx=k$$ then by partial fractions we get:
$$\lim_{N \to \infty} \int_{3}^N \left(\frac{2}{3}\frac{1}{x}+\frac{a-\frac{2}{3}}{x+3}-\frac{b}{3x-2}\right) dx=k$$
then
$$\lim_{N \to \infty} \left[\frac{2}{3}ln(x)+(a-\frac{2}{3})ln (x+3)-\frac{b}{3}ln(3x-2)\right]_3^N =k$$
By logarithm properties:
$$\lim_{N \to \infty} \left[ln\frac{(x)^\frac{2}{3}(x+3)^\left(a-\frac{2}{3}\right)}{(3x-2)\frac{b}{3}}\right]_3^N =k$$
Evaluating:
$$\lim_{N \to \infty} \left[ln\frac{(N)^\frac{2}{3}(N+3)^\left(a-\frac{2}{3}\right)}{(3N-2)^\frac{b}{3}}\right]-\left[ln\frac{(3)^\frac{2}{3}(6)^\left(a-\frac{2}{3}\right)}{(7)^\frac{b}{3}}\right] =k$$
But, I do not know what to do after this step. Any recommendation will be appreciated

Hint. From your third line, one may write, as $x \to \infty$,
$$
\begin{align}
&\color{blue}{\frac{2}{3}}\ln(x)+\color{blue}{\left(a-\frac{2}{3}\right)}\ln (x+3)\color{blue}{-\frac{b}{3}}\ln(3x-2)
\\=&\color{blue}{\left(\frac{2}{3}+a-\frac{2}{3}-\frac{b}{3}\right)}\ln x+\left(a-\frac{2}{3}\right)\ln \left(1+\frac 3x\right)-\frac{b}{3}\cdot\ln\left(3-\frac 2x\right)
\\=&\color{blue}{\left(a-\frac{b}{3}\right)}\ln x+O(1)
\end{align}
$$ then all one needs is to annihilate the factor of $\ln x$, that is taking
$$
\color{blue}{a=\frac b3}
$$ in order to guarantee the convergence of the given integral.
