<s>[INST] How do I find the closed form of this integral $\int_0^2\frac{\ln x}{x^3-2x+4}dx$? How do I find the closed form of this integral:
$$I=\int_0^2\frac{\ln x}{x^3-2x+4}dx$$
First, I have a partial fraction of it:
$$\frac{1}{x^3-2x+4}=\frac{1}{(x+2)(x^2-2x+2)}=\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+2}$$
$$A=\frac{1}{(x^3-2x+4)'}|_{x=-2}=\frac{1}{(3x^2-2)}|_{x=-2}=\frac{1}{10}$$
$$Bx+C=\frac{1}{x+2}|_{x^2-2x=-2}=\frac{1}{(x+2)}\frac{(x-4)}{(x-4)}|_{x^2-2x=-2}=$$
$$=\frac{(x-4)}{(x^2-2x-8)}|_{x^2-2x=-2}=\frac{(x-4)}{(-2-8)}|_{x^2-2x=-2}=-\frac{1}{10}(x-4)$$
Thus:
$$\frac{1}{x^3-2x+4}=\frac{1}{10}\left(\frac{1}{x+2}-\frac{x-4}{x^2-2x+2}\right)$$
$$I=\frac{1}{10}\left(\int_0^2\frac{\ln x}{x+2}dx-\int_0^2\frac{(x-4)\ln x}{x^2-2x+2}dx\right)$$
What should I do next?
 [/INST] Just for your curiosity.
If you enjoy special functions of complex arguments, the antiderivative can be computed.
$$\frac 1 {x^3-2x+4}=\frac{\frac1{10}}{
   x+2}-\frac{\frac{1}{20}-\frac{3 i}{20}}{x-(1-i)}-\frac{\frac{1}{20}+\frac{3 i}{20}}{x-(1+i)}$$ which makes that we are left with integrals $$I_a=\int \frac{\log(x)}{x-a}\,dx=\text{Li}_2\left(\frac{x}{a}\right)+\log (x) \log \left(1-\frac{x}{a}\right)$$ where appears the polylogarithm function.
This would lead to the result  Thomas Andrews gave in a comment.
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