Answer
$\frac{1}{2}\left(\frac{1}{2}\ln|x^2+3|-\frac{1}{\sqrt 3}\tan^{-1}\left(\frac{x}{\sqrt 3}\right)+\frac{1}{2}\ln|x^2+1|-3\tan^{-1}x\right)+C$
Work Step by Step
$\int\frac{x^{3}-2x^{2}+2x-5}{x^{4}+4x^{2}+3}dx$ = $\int\left[\frac{x^{3}-2x^{2}+2x-5}{(x^{2}+3)(x^{2}+1)}\right]dx$ = $\int\left(\frac{Ax+B}{x^{2}+3}+\frac{Cx+D}{x^{2}+1}\right)dx$
$x^{3}-2x^{2}+2x-5$ = $(Ax+B)(x^{2}+1)+(Cx+D)(x^{2}+3)$
$x^{3}-2x^{2}+2x-5$ = $(Ax^3+Bx^2+Ax+B)+(Cx^3+Dx^2+3Cx+3D)$
$1$ = $A+C$
$-2$ = $B+D$
$2$ = $A+3C$
$-5$ = $B+3D$
$A$ = $\frac{1}{2}$, $B$ = $-\frac{1}{2}$, $C$ = $\frac{1}{2}$, $D$ = $-\frac{3}{2}$
So
$\int\frac{x^{3}-2x^{2}+2x-5}{x^{4}+4x^{2}+3}dx$ = $\int{\frac{1}{2}}\left(\frac{x-1}{x^{2}+3}+\frac{x-3}{x^{2}+1}\right)dx$ = $\int{\frac{1}{2}}\left[\frac{x}{x^{2}+3}-\frac{1}{x^{2}+3}+\frac{x}{x^{2}+1}-\frac{3}{x^{2}+1}\right]dx$ = $\frac{1}{2}\left[\frac{1}{2}\ln|x^2+3|-\frac{1}{\sqrt 3}\tan^{-1}\left(\frac{x}{\sqrt 3}\right)+\frac{1}{2}\ln|x^2+1|-3\tan^{-1}x\right]+C$
