Integrate: $\int \frac{2x^2-3x+8}{x^3+4x} \, dx$ $$\int \frac{2x^2-3x+8}{x^3+4x}\,dx$$
My main problem is calculating the $B$ and $C$. This is the algebra part. Thus, what is a technique I can use that is in line with what I did to calculate $A$? 

Since you're going to set $x=2t$ anyway, because of the factor $x^2+4$, do it at the start: the integral becomes
$$
\int\frac{8t^2-6t+8}{4t(t^2+1)}\,2dt
=
\int\frac{4t^2-3t+4}{t(t^2+1)}\,dt
=
\int\left(\frac{4}{t}-\frac{3}{t^2+1}\right)\,dt=4\log|t|-3\arctan t+c
$$
So your original integral is
$$
\int\frac{2x^2-3x+8}{x^3+4x}\,dx=4\log|x|-3\arctan\frac{x}{2}+c
$$
