Answer
$$-\ln x+\frac{1}{2}\ln |x^2+3| +\frac{2}{\sqrt{3}} \tan^{-1}\frac{x}{\sqrt{3}}+c$$
Work Step by Step
Given $$\int\frac{2x-3}{x^3+3x}dx$$
Since
\begin{align*}
\frac{2x-3}{x^3+3x}&=\frac{A}{x }+\frac{Bx+C}{x^2+3 }\\
&=\frac{Bx^2+Cx+Ax^2+3A}{x^2+3 }\\
2x-3&= Bx^2+Cx+Ax^2+3A
\end{align*}
At $x=0\to A=-1$ and $ A+B=0\to B=1$ , $C=2$ , then
\begin{align*}
\int\frac{2x-3}{x^3+3x}dx&=\int\frac{-1}{x }dx+\int\frac{ x+2}{x^2+3 }dx\\
&=\int\frac{-1}{x }dx+\int\frac{ x }{x^2+3 }dx+\int\frac{ 2}{x^2+3 }dx\\
&=-\ln x+\frac{1}{2}\ln |x^2+3| +\frac{2}{\sqrt{3}} \tan^{-1}\frac{x}{\sqrt{3}}+c
\end{align*}
