Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 1 down vote accepted

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

share|cite|improve this answer
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


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.