integration by partial fraction I can not figure out how to integrate 
$$\int\frac{3x^2+x+4}{x^3+x} \, dx.$$
I got as far as factoring the denominator which gives us $x(x^2+1)$.
So from there I got
$$\frac{3x^2+x+4}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$$
which would give us: $(3x^2 + x + 4) = A(x^2 + 1) + (Bx+C)x$
This is where Im stuck because I can not figure out the values for $B$ and $C$. Please help.

Hint:
In the equation $(3x^2 + x + 4) = A(x^2 + 1) + (Bx+C)x$, by matching coefficients of the quadratic term $x^2$ we find:
$$A+B=3,$$
by matching coefficients of the linear term $x$ we find:
$$C=1,$$
and lastly, by matching the constant term we find:
$$A=4$$ 
Can you solve it from here?
