<s>[INST] 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
 [/INST] $$\left(\dfrac{1}{2}\ln (x-3)- \dfrac{1}{3}\ln (x)- \dfrac{1}{6}\ln (x+3)+C\right)'=\frac1{2(x-3)}-\frac1{3x}-\frac1{6(x+3)}
\\=\frac{18(x^2+3x)-12(x^2-9)-6(x^2-3x)}{12x(x^2-9)}
\\=\frac{72x+108}{36(x^3-9x)}.$$
It seems that you are right.
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