For every quadratic factor $x^2 + bx + c$, we have $$\displaystyle\frac{Ax+B}{x^2+bx+c}.$$

For every repeated quadratic factor $(x^2+bx+c)^n$, we have $$\displaystyle\frac{A_1 x + B_1}{x^2+bx+c} + \frac{A_2x+B_2}{(x^2+bx+c)^2}+\ldots + \frac{A_nx+B_n}{(x^2+bx+c)^n}.$$

Example: We have the decomposition
$$
\displaystyle{ 2x^3+5x-1 \over (x+1)^3(x^2+1)^2 } = \displaystyle{ { A \over x+1 } + { B \over (x+1)^2 }+ { C \over (x+1)^3 } + { Dx+E \over x^2+1 } + { Fx+G \over (x^2+1)^2 } }. $$

Finding the coefficients $A$, $B$, etc. is also harder. After cross-multiplying, we can't just plug in the roots of $Q(x)$ to get the coefficients one at a time, since the quadratic factors don't have real roots. We either have to plug in lots of different values of $x$, or compare the corresponding coefficients of $1$, $x$, $x^2$, etc. Either way, we get a system of linear equations to solve.

Finally, solving the integrals at the end is harder than with linear factors.

To integrate $$\displaystyle\int \frac{x}{(x^2+a^2)^n}\,dx,$$ we can substitute $u=x^2+a^2$.

To integrate $$\displaystyle\int \frac{dx}{(x^2+a^2)^n},$$
on the other hand, we can substitute $x=a\tan(\theta)$.

These complications, and how to overcome them, are explained in the following video.