Integral of $\int\frac{(x^4+1)\,dx}{x^3+4x}$ I followed the steps to solve this integral and want to know if I did it right and if $C=0? $ 
$$\int\frac{(x^4+1)\,dx}{x^3+4x} = \int\frac{(x^4+1)\,dx}{x(x^2+4)} = \frac{A}{x}+\frac{Bx+C}{x^2+4}$$
 $$(x^2+4)A+x(Bx+C)=x^4+1$$
 $$x=0 => 4A=1 => A=\frac{1}{4}$$
 $$Ax^2+4A+Bx^2+Cx=x^4+1 = > (A+B)x^2+4A+Cx=x^4+1$$
 $$A+B=0 => B=-\frac{1}{4}, C=0$$
Thanks!

Whatever $A$, $B$, $C$ may be, the numerator of $\frac{A}{x} + \frac{B x+C}{x^2+4}$ is no greater than 3. The proper ansatz is
$$
   \frac{x^4+1}{x^3+4x} = x + \frac{1-4x^2}{x(x^2+4)} = x + \frac{A}{x} + \frac{B x+ C}{x^2+4}
$$
and you should get $A = \frac{1}{4}$ and $C=0$, $B = -\frac{17}{4}$.
