# Computing $\int \dfrac{x^4}{x^3-8} dx$

I am currently stuck with one integration and I don't know what to do know.

I would appreciate detailed answer, but will be happy for any :-)

I am having: $$\int \dfrac{x^4}{x^3-8} \,dx$$ I am having 2 complex roots and I am not able to find any "clever" substitution for this. I have google a lot and I would be really grateful if you can tell me about some e-books/books about similar kind of integrations.

Best thanks.

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Hint: $$\int \dfrac{x^4}{x^3-8} \,dx=\int \dfrac{x^4-8x+8x}{x^3-8} \,dx \\ =\int x+\dfrac{8x}{x^3-8} \,dx=\int x+\dfrac{8x}{(x-2)(x^2+2x+4)} \,dx$$ where $$\dfrac{8x}{(x-2)(x^2+2x+4)}=\dfrac{A}{(x-2)}+\dfrac{Bx+C}{(x^2+2x+4)}$$