<s>[INST] Integral fraction of polynomials I have this problem:
$$\int \frac{-2x^2+6x+8}{x^4-4x+3}dx$$
I have tried using partial fractions, but I can't get solution. Thank you for any advice.
 [/INST] $x=1$ is a double root of $x^4-4x+3$ (it is also a root of the derivative), hence it is divisible by $(x-1)^2$, hence you'll get $x^4-4x+3=(x-1)^2(x^2+ax+b)$, and partial fractions decomposition has the form:
$$\frac{-2x^2+6x+8}{x^4-4x+3}=\frac A{x-1}+\frac B{(x-1)^2}+\frac{Cx+D}{x^2+ax+b}.$$
You'll obtain $A\ln\lvert x-1\rvert-\dfrac B{x-1} +C'\ln \lvert x^2+ax+b\rvert+D'\arctan(\text{something})$.
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