Answer
$\frac{1}{2}\ln|x^2+1|+\tan^{-1}x-\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)+C$
Work Step by Step
$\int\frac{x^{3}+4x+3}{x^{4}+5x^{2}+4}dx$ = $\int\frac{x^{3}+4x+3}{(x^2+4)(x^2+1)}dx$ = $\int\left(\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4}\right)dx$
$x^{3}+4x+3$ = $(Ax+B)(x^2+4)+(Cx+D)(x^2+1)$
$x^{3}+4x+3$ = $(Ax^3+Bx^2+4Ax+4B)+(Cx^3+Dx^2+Cx+D)$
$1$ = $A+C$
$0$ = $B+D$
$4$ = $4A+C$
$3$ = $4B+D$
$A$ = $1$, $B$ = $1$, $C$ = $0$, $D$ = $-1$
So
$\int\frac{x^{3}+4x+3}{x^{4}+5x^{2}+4}dx$ = $\int\left(\frac{x+1}{x^2+1}-\frac{1}{x^2+4}\right)dx$ = $\int\left(\frac{x}{x^2+1}+\frac{1}{x^2+1}-\frac{1}{x^2+4}\right)dx$ = $\frac{1}{2}\ln|x^2+1|+\tan^{-1}x-\frac{1}{2}\tan^{-1}(\frac{x}{2})+C$
