Evaluating $\int\frac{4x^3-3x^2+6x-27}{x^4+9x^2}dx$

Question

$$\int\frac{4x^3-3x^2+6x-27}{x^4+9x^2}dx$$

this integral get very messy. Can I get a step by step breakdown of solving?

Answer 1

No Step by step solution, but just a hint

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

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

Answer 2

Doing long division, I get $$\int{\frac{4}{x}-\frac{3}{x^2}-\frac{30}{x^3+9x}dx}$$ We can rewrite as $$\int{4x^{-1}-3x^{-2}-30(x^3+9x)^{-1}}$$