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.

  • 2
    $\begingroup$ What have you tried to do in order to find $A$, $B$ and $C$? The procedure to compute the partial fraction decomposition is quite the same every time... $\endgroup$ Mar 10, 2011 at 1:48
  • $\begingroup$ (Also, in the second equation the integral sign should not appear) $\endgroup$ Mar 10, 2011 at 1:52
  • 2
    $\begingroup$ See the Heaviside cover-up method and see this prior question $\endgroup$ Mar 10, 2011 at 1:55

3 Answers 3


Okay, first a bunch of nitpicks:

  • You are not trying to integrate the integral, you are trying to integrate the function itself.

  • The integral does not equal the partial fraction decomposition, only the rational function itself does.

  • You are missing the $dx$ in all the integrals.

So. You are trying to integrate $\displaystyle \frac{3x^2+x+4}{x^3+x}$.

First, you make sure the numerator has smaller degree than the denominator, doing long division if necessary to get it into that form (done).

Then, you factor the denominator completely (done).

Then you set up the partial fraction decomposition problem (done): $$\frac{3x^2+x+4}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}.$$

Then, you perform the operation on the right hand side to the expression you want (done): $$\frac{3x^2+x+4}{x(x^2+1)} = \frac{A(x^2+1) + (Bx+C)x}{x(x^2+1)}.$$

Then, you know the numerators are equal (done): $$3x^2 + x+4 = A(x^2+1)+ (Bx+c)x.$$

Finally, you figure out the values of $A$, $B$, and $C$. There are two strategies:

  1. Do the operations on the right and write it as a polynomial; since two polynomials are equal if and only if they have the same coefficients, the coefficient of $x^2$ on the right equals $3$, etc. This will set up a system of linear equations for $A$, $B$, and $C$, which you can solve: $$3x^2 + x + 4 = Ax^2 + A + Bx^2 + Cx = (A+B)x^2 + Cx + A$$ so $A+B=3$, $C=1$, and $A=4$. From this, you get $A=4$, $B=-1$, and $C=1$, so $$\frac{3x^2+x+4}{x(x+1)} = \frac{4}{x} + \frac{-x+1}{x^2+1}.$$ Now you just need to do the simpler integrals $$\int\frac{3x^2+x+4}{x^3+x}\,dx = \int\frac{4}{x}\,dx - \int\frac{x}{x^2+1}\,dx + \int\frac{1}{x^2+1}\,dx.$$

  2. Plug in some values of $x$ to get information about $A$, $B$, and $C$. Specifically, pick values that make some of the terms equal to $0$ (the roots of the original polynomial), and start simplifying. For example, from $$3x^2 + x + 4 = A(x^2+1) + (Bx+C)x,$$ plugging in $x=0$ you get $4 = A(1) + 0$, so $A=4$; now we have $$3x^2 + x + 4 = 4x^2 + 4 + (Bx+C)x.$$ Moving all the known factors to the left, we have $$-x^2 + x = (Bx+C)x$$ and factoring out $x$, we get $x(-x+1) = (Bx+C)x$, from which you can cancel $x$ to get $$-x+1 = Bx+C$$ which immediately gives $B=-1$ and $C=1$, as before. Now that we know $A$, $B$, and $C$, proceed as in 1.

  • 5
    $\begingroup$ A simpler way to find the value of $\rm\:b,c\:$ is to evaluate $\rm\ 3\ x^2 + x + 4 = a\ (x^2+1) + (b\ x +c)\ x\ $ at a root of $\rm\:x^2+1\:,\:$ say at $\rm\:x = i\:.\ $ This yields $\rm\ i+1 = c\ i - b\ $ so $\rm b = -1,\ c = 1\:.\:$ This is a special case of the higher-degree Heaviside cover-up method that I mentioned here. $\endgroup$ Mar 10, 2011 at 3:47
  • $\begingroup$ @Bill: We need not even multiply through to use the cover method you mention. Then it just becomes a statement about the asymptotics of functions. I do agree, what you said is fastest for beginner calculus students. But the nicest (and easiest for me) is to notice that partial fractions really live in the complex numbers. Finding the coefficients is the same as finding the residues at the poles. $\endgroup$ Mar 10, 2011 at 4:35
  • $\begingroup$ @Eric: Yes, the arithmetic is transmuted into "covering-up" operations - hence the name. I do the arithmetic in my linked post merely to show why covering-up works. One can of course use residue calculus or analogous techniques. But they would be overkill here. $\endgroup$ Mar 10, 2011 at 4:38


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?

  • $\begingroup$ Yeah i see how you got that. My professor just taught us to plug in numbers for x and solve for the variables. i.e x=0 then A=4 because B and C cancel, which is also what you got above. but i couldnt find anything to cancel the A because of the x^2. Thank you for all your help $\endgroup$
    – user8051
    Mar 10, 2011 at 2:04
  • 3
    $\begingroup$ If two functions are equal, they have to be equal for any x, so you can choose any three values of x to plug in to find A,B, and C. It just happens that choosing values that will cancel out some of the terms (if there are any) are the easiest ones to choose. If there aren't enough like this you can choose any other "easy" values. Eric's method will always work too. $\endgroup$
    – Brian
    Mar 10, 2011 at 2:36
  • 1
    $\begingroup$ @user8051: After you find that $A=4$, plug it in and simplify, like I did in my answer above. $\endgroup$ Mar 10, 2011 at 2:55

Observe that:

$$ \begin{align}\frac{3x^2+x+4}{x^3+x}=\frac{3(x^2+1)-2x+1}{x(x^2+1)}&=\frac{3}{x}-\frac{2}{x^2+1}+\frac{1}{x}-\frac{x}{x^2+1}\\ &=\frac{4}{x}-\frac{2+x}{x^2+1} \end{align}$$

and the rest should be easy.


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