Answer
\[\int \frac{x^3+4x+3}{x^4+5x^2+4} dx =\frac{1}{2}\ln |x^2+1|+\arctan (x)-\frac{1}{2}\arctan (\frac{x}{2})+C\]
Work Step by Step
Make \[\frac{x^3+4x+3}{x^4+5x^2+4}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4} = \frac{(A+C)x^3+(B+D)x^2+(4A+C)x+4B+D}{x^4+5x^2+4} \]
$\therefore$ \[
\begin{cases}
A=1\\
B=1\\
C=0\\
D=-1\\
\end{cases} \]
$\therefore$\[\int \frac{x^3+4x+3}{x^4+5x^2+4} dx = \int (\frac{x+1}{x^2+1}+\frac{-1}{x^2+4}) dx=\frac{1}{2}\ln |x^2+1|+\arctan (x)-\frac{1}{2}\arctan (\frac{x}{2})+C\]
