# 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!

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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 $C=0$, $B = -\frac{17}{4}$.

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The degree of the polynomial in the numerator can be no greater than 3. You can actually do the division to get the x term, then the remainder can be split with partial fraction decomposition as shown above in this answer. – agktmte May 7 '13 at 14:33
Why its $-\frac{17}{4}$? – Ofir Attia May 7 '13 at 14:36

There are no constants $A,B,C$ such that

$$\frac{x^4+1}{x^3+4x} = \frac{A}{x}+\frac{Bx+C}{x^2+4},$$

because the integrand is a rational function in $x$ and the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator. So prior to expanding it into partial fractions, the standard technique is to rewrite it as

$$\frac{x^4+1}{x^3+4x}=x+\frac{1-4x^2}{x^3+4x}$$

by using polynomial long division or Ruffini's rule.

Now you can proceed by expanding $\frac{1-4x^2}{x^3+4x}$ into partial fractions.

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