how can I integrate $\int \dfrac{2x+3}{x^3-9x}dx$? How can I integrate this $$\int \dfrac{2x+3}{x^3-9x}dx?$$
My work
$$\dfrac{2x+3}{x^3-9x}=\dfrac{2x+3}{x(x+3)(x-3)}$$
$$=\dfrac{A}{x-3}+\dfrac{B}{x}=\dfrac{C}{x+3}$$
after solving, I got $A=1/2$, $B=-1/3$, $C=-1/6$
partial fractions decompositions:
$$\dfrac{2x+3}{x^3-9x}=\dfrac{1/2}{x-3}+\dfrac{-1/3}{x}=\dfrac{-1/6}{x+3}$$
$$\int \left(\dfrac{1}{2(x-3)}- \dfrac{1}{3x}- \dfrac{1}{6(x+3)}\right)\ dx$$
$$=\dfrac{1}{2}\ln (x-3)- \dfrac{1}{3}\ln (x)- \dfrac{1}{6}\ln (x+3)+C$$
I am not sure if my answer is correct. my question is can I use substitution to solve above integration? if yes please help me solve it by substitution. thanks

Your working is correct.  Use absolute values for logarithms in final answer
For substitution, let $x=3\sec\theta\implies dx=3\sec\theta\tan\theta d\ \theta $
$$\int \dfrac{2x+3}{x^3-9x}dx=\int \frac{2\sec\theta+3}{3\sec\theta(9\tan^2\theta)}3\sec\theta\tan\theta d\ \theta $$
$$=\frac19\int \frac{2\sec\theta+3}{\tan\theta} d\theta$$
$$=\frac19\int (2\csc\theta+3\cot\theta) d\theta$$
$$=\frac29\int \csc\theta d\theta+\frac13\int \cot\theta d\theta$$
$$=\frac29\ln\left|\csc\theta-\cot\theta\right|+\frac13\ln|\sin\theta|+C$$
Substituting back to $x$,
$$=\frac1{9}\ln\left|\frac{x-3}{x+3}\right|+\frac16\ln\left|\frac{x^2-9}{x^2}\right|+C$$
