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.

A quick check shows that $x = 1$ is a root of $x^4 - 4x + 3$. By synthetic division we find $x^4 - 4x + 3 = (x - 1)(x^3 + x^2 + x - 3)$. By similar reasoning, $x^3 + x^2 + x - 3 = (x - 1)(x^2 + 2x + 3)$. So $x^4 - 4x + 3 = (x - 1)^2(x^2 + 2x + 3)$. Now you have the setup to do partial fraction decomposition. The result is  
$$\frac{-2x^2 + 6x + 8}{x^4 - 4x + 3} = \frac{2}{(x - 1)^2} - \frac{1}{x - 1} + \frac{x - 1}{x^2 + 2x + 3}.$$
So 
\begin{align}
\int \frac{-2x^2 + 6x + 8}{x^4 - 4x + 3}\, dx &= \int \frac{2\, dx}{(x - 1)^2} - \int \frac{dx}{x - 1} + \int \frac{x - 1}{x^2 + 2x + 3}\, dx\\
&= -\frac{2}{x - 1} - \ln|x - 1| + \int \frac{x - 1}{(x + 1)^2 + 2}\, dx\\
&= -\frac{2}{x - 1} - \ln|x - 1| + \int \frac{u - 2}{u^2 + 2}\, du \quad (u = x + 1)\\
&= - \frac{2}{x - 1} - \ln|x - 1| + \int \frac{u}{u^2 + 2} - \int \frac{2}{u^2 + 2}\, du\\
&= -\frac{2}{x - 1} - \ln|x - 1| + \frac{1}{2}\ln|u^2 + 2| - \frac{2}{\sqrt{2}}\arctan \frac{u}{\sqrt{2}} + C\\
&= -\frac{2}{x - 1} - \ln|x - 1| + \frac{1}{2}\ln[(x + 1)^2 + 2] - \sqrt{2}\arctan \frac{x+1}{\sqrt{2}} + C.
\end{align}
