<s>[INST] Evaluating $\int\frac{4x^3-3x^2+6x-27}{x^4+9x^2}dx$ $$\int\frac{4x^3-3x^2+6x-27}{x^4+9x^2}dx$$
this integral get very messy. Can I get a step by step breakdown of solving? 
 [/INST] 
No Step by step solution, but just a hint

By reducing to partial fractions we get
$$\frac{4x^3-3x^2+6x-27}{x^4+9x^2}=\frac{Ax+ B}{x^2+9}+\frac{C}{x^2}+\frac Dx$$
$$Ax^3+Bx^2+Cx^2+9C+Dx^3+9Dx=4x^3-3x^2+6x-27$$
$$9C=-27\implies C=-3$$
$$9D=6\implies D=\frac23$$
$$A+D=4\implies A=-\frac{10}{3}$$
$$B+C=-3\implies B=0$$

$$\frac{4x^3-3x^2+6x-27}{x^4+9x^2} =\frac{10x}{3(x^2+9)}-\frac{3}{x^2}+\frac{2}{3x}$$

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