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 $A$? 
 A: You can equate powers of $x$. So you see that your equation simplifies to $$2x^2 -3x+8=Ax^2+4A+Bx^2+Cx$$
If we equate all powers of $x$ we get the system of equations
$$x^2:2=A+B$$
$$x:-3=C$$
$$x^0:8=4A$$
From here it is clear how to get $B,C$.
A: Continuing from your work,
$$2x^2-3x+8=A(x^2+4)+(Bx+C)x$$
Now $x=0$ gives $A=2$
Also we can write the above equation as :
$$2x^2-3x+8=(A+B)x^2+4A+Cx$$
This is an identity in $x$.
So coefficients of the respective powers on both sides must be same.
$$A+B=2$$ and $$C=-3$$
So $$B=0$$
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
$$
\int\frac{2x^2-3x+8}{x^3+4x}\,dx=4\log|x|-3\arctan\frac{x}{2}+c
$$
A: 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
$$B=0,\enspace C=-3.$$

