Answer
$\displaystyle \frac{1}{4}\ln|x^{2}+1|-\frac{3}{2}\arctan(x))+\frac{1}{4}\ln|x^{2}+3|-\frac{1}{2\sqrt{3}}\arctan(\frac{x}{\sqrt{3}}))+C$
Work Step by Step
$x^{4}+4x^{2}+3$ can be factored:
1 and 3 multiplied give 3, and their sum is 4....
$\displaystyle \frac{x^{3}-2x^{2}+2x-5}{(x^{2}+1)(x^{2}+3)}=\frac{Ax+B}{x^{2}+1}+\frac{Cx+D}{x^{2}+3}$
$x^{3}-2x^{2}+2x-5=(Ax+B)(x^{2}+3)+(Cx+D)(x^{2}+1)$
$x^{3}-2x^{2}+2x-5=Ax^{3}+3Ax+Bx^{2}+3B+ Cx^{3}+Cx+Dx^{2}+D$
$x^{3}-2x^{2}+2x-5=(A+C)x^{3}+(B+D)x^{2}+(3A+C)x+(3B+D)$
$\left\{\begin{array}{l}
A+C=1\\
3A+C=2
\end{array}\right\}$, subtracting, $2A=1\displaystyle \Rightarrow A=\frac{1}{2},C=\frac{1}{2}$
$\left\{\begin{array}{l}
B+D=-2\\
3B+D=-5
\end{array}\right\}$, subtracting, $2B=-3\displaystyle \Rightarrow B=\frac{-3}{2},D=\frac{-1}{2}$
$\displaystyle \frac{x^{3}-2x^{2}+2x-5}{(x^{2}+1)(x^{2}+3)}=\frac{\frac{1}{2}x+\left(-\frac{3}{2}\right)}{x^{2}+1}+\frac{\frac{1}{2}x+\left(-\frac{1}{2}\right)}{x^{2}+3}$
$=\displaystyle \frac{x-3}{2(x^{2}+1)}+\frac{x-1}{2(x^{2}+3)}$
$\displaystyle \frac{x-3}{2(x^{2}+1)}=\frac{1}{4}\cdot\frac{2x}{(x^{2}+1)}-\frac{3}{2}\cdot\frac{1}{(x^{2}+1)}$
$\displaystyle \frac{x-1}{2(x^{2}+3)}=\frac{1}{4}\cdot\frac{2x}{(x^{2}+3)}-\frac{1}{2}\cdot\frac{1}{x^{2}+(\sqrt{3})^{2}}$
$\displaystyle \int\frac{x^{3}-2x^{2}+2x-5}{x^{4}+4x^{2}+3}dx= $
$=\displaystyle \frac{1}{4}\ln|x^{2}+1|-\frac{3}{2}\arctan(x))+\frac{1}{4}\ln|x^{2}+3|-\frac{1}{2\sqrt{3}}\arctan(\frac{x}{\sqrt{3}}))+C$
