integrate $\int \frac{x^4+4x+6}{(2-x)^2(4+x^2)}$ 
$$\int \frac{x^4+4x+6}{(2-x)^2(4+x^2)}$$

$(2-x)^2(4+x^2)=x^4-4x^3+8x^2-16x+16$
$$\int \frac{x^4+4x+6}{x^4-4x^3+8x^2-16x+16}$$
Is there is anything to do else than long division?

I know that you asked without long division, but
$$\int \frac{x^4+4x+6}{(x-2)^2(x^2+4)}\stackrel{\text{L.div}}{=}\frac 1 8 \int\frac{11x-8}{x^2+4}dx+\frac{21}{8}\int\frac{dx}{x-2}dx+\frac{15}{4}\int \frac{dx}{(x-2)^2}+\int1 dx=\dots$$
