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: 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}
A: As one can obviously see, $x=1$ is a root of the denominator, and thus, it can be divided by $(x-1)$. Using long division, we get $(x-1)(x^3+x^2+x-3)$
Again, we can see that $x=1$ is a solution, of the second factor, so, using long division again, we get $(x-1)^2(x^2+2x+3)$
$$\frac{-2x^2+6x+8}{(x-1)^2(x^2+2x+3)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Ex+F}{(x^2+2x+3)}$$
Thus, we have 
$$-2x^2+6x+8=A(x-1)(x^2+2x+3)+B(x^2+2x+3)+(Ex+F)(x-1)^2$$
$$-2x^2+6x+8=A(x^3+2x^2+3x-x^2-2x-3)+Bx^2+2Bx+3B+(Ex+F)(x^2-2x+1)$$
By identification, we find 
$A=-1$, $B=2$ $E=1$ $F=-1$
$$\int \frac{-1}{x-1}\,dx =-\ln(x-1)$$
$$\int \frac{2}{(x-1)^2} \,dx =\int 2(x-1)^{-2} =-2(x-1)^{-1}\,dx$$
$$\int \frac{x-1}{x^2+2x+3}\,dx=\int \frac{x+1}{x^2+2x+3}\,dx -2\int \frac{1}{x^2+2x+3}\,dx =\frac{1}{2}\ln(x^2+2x+3)-2\int \frac{1}{2+(x+1)^2} \,dx =\frac{1}{2}\ln(x^2+2x+3)-\int \frac{1}{1+(\frac{x+1}{\sqrt{2}})^2} \,dx =\frac{1}{2}\ln(x^2+2x+3)-\sqrt{2}\arctan \frac{x+1}{\sqrt{2}}$$
Not sure if I made any mistake since it's pretty late and I'm dead tired, but I hope this gets the point. 
A: $x=1$ 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 $A\ln\lvert x-1\rvert-\dfrac B{x-1} +C'\ln \lvert x^2+ax+b\rvert+D'\arctan(\text{something})$.
