Answer
$ -\displaystyle \ln x+\frac{1}{2}\ln(x^{2}+3)+\frac{2}{\sqrt{3}}\tan^{-1}(\frac{x}{\sqrt{3}})+C$
Work Step by Step
Strategy: Rational function. Partial fractions
$\displaystyle \frac{2x-3}{x(x^{2}+3)}=\frac{A}{x}+\frac{Bx+C}{x^{2}+3}$
$\displaystyle \frac{2x-3}{x(x^{2}+3)}=\frac{Ax^{2}+3A+Bx^{2}+Cx}{x(x^{2}+3)}$
$2x-3=(A+B)x^{2}+Cx+3A\Rightarrow\left\{\begin{array}{l}
A+B=0\\
C=2\\
3A=-3
\end{array}\right.$
$A=-1, B=1, C=2$
$\displaystyle \int\frac{2x-3}{x(x^{2}+3)}dx=\int-\frac{1}{x}dx+\int\frac{x+2}{x^{2}+3}$
$=-\displaystyle \int\frac{1}{x}dx+\int\frac{xdx}{x^{2}+3}+2\int\frac{dx}{x^{2}+3}=-I_{1}+I_{2}+2I_{3}$
$I_{1}=\ln x+C_{1}$
$I_{2}=\displaystyle \left[\begin{array}{l}
t=x^{2}+3\\
dt=2xdx
\end{array}\right]=\frac{1}{2}\int\frac{dt}{t}=\frac{1}{2}\ln t=\frac{1}{2}\ln(x^{2}+3)+C_{2}$
$I_{3}=\displaystyle \left[17. \quad \int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a}\right]=\frac{1}{\sqrt{3}}\tan^{-1}(\frac{x}{\sqrt{3}})+C_{3}$
$-I_{1}+I_{2}+2I_{3}=-\displaystyle \ln x+\frac{1}{2}\ln(x^{2}+3)+\frac{2}{\sqrt{3}}\tan^{-1}(\frac{x}{\sqrt{3}})+C$
