<s>[INST] 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 $A$? 
 [/INST] Multiplying both sides of $$\frac{2x^2-3x+8}{x^3+4x}=\frac Ax+\frac{Bx+C}{x^2+4}$$
with $x(x^2+4)$, you can remove denominators, and get
$$2x^2-3x+8=A(x^2+4)+(Bx+C)x. $$
Then set $x=\;$ the poles of the fraction:


*

*$x=0$ yields $8=4A+0$, whence  $A=2$.

*$x=2\mathrm i$  yields $-8-6\mathrm i+8=-6\mathrm i=0+(2B\mathrm i+C)2\mathrm i=-4B+2C\mathrm i$. Identifying the real and imaginary parts, we get
$$B=0,\enspace C=-3.$$

</s>