# Why is numerator re-defining necessary for proper solution of some partial fraction integrations?

In my homework, one problem was the following formula. Using standard partial fraction techniques where you'll see my work, I came up with an almost correct answer in the fact the book solution had a three term integral rather than the two term I used. The only difference between my work and the correct solution was I didn't rewrite the numerator. My two-part question is

• Why did excluding the numerator re-defining result in the wrong answer?
• Under what circumstances should I use numerator re-defining?

Given the following formula, integrate using partial fractions:

$$\int\frac{2x^3-4x^2-15x+5}{x^2-2x-8}$$

\begin{align*} 2x^3-4x^2-15x+5=\frac{A}{x-4}+\frac{B}{x+2}\\ 2x^3-4x^2-15x+5=A(x+2)+B(x-4)\\ \end{align*}

Let $Ax$=-2

\begin{align*} 2x^3-4x^2-15x+5=A(x+2)+B(x-4)\\ 2(-2)^3-4(-2)^2-15(-2)+5=A(-2+2)+B(-2-4)\\ 3=-6B\\ B=-\frac{1}{2} \end{align*}

Let $Bx= 4$

\begin{align*} 2x^3-4x^2-15x+5=A(x+2)+B(x-4)\\ 2(4)^3-4(4)^2-15(4)+5=A(4+2)+B(4-4)\\ 9=6A\\ A=\frac{3}{2} \end{align*}

Using $A \text{ and } B$ values, plug them into the integral as

\begin{align*} \frac{3}{2}\int\frac{1}{x-4}\text{dx}-\frac{1}{2}\int\frac{1}{x+2}\text{dx}\\ =\frac{3}{2}\text{ln}\left |x-4 \right |-\frac{1}{2}\text{ln}\left |x+2\right |+C \end{align*}

This work matches up with the solution process except the original integral was re-written as

$$\int2x+\frac{x+5}{(x-4)(x+2)}$$

which resulted in the final integral terms

$$\int2x\text{dx}+\frac{3}{2}\int\frac{1}{x-4}\text{dx}-\frac{1}{2}\int\frac{1}{x+2}\text{dx}$$

• You're missing terms if the numerator has bigger degree than the denominator. Either you need to add a linear and constant term, or you need to allow $A$ and $B$ to be polynomials instead of constants. – Qiaochu Yuan Aug 11 '11 at 22:00
• You should do long division first. – The Chaz 2.0 Aug 11 '11 at 22:18
• Your second equation is nonsense. Your third equation doesn't follow from the second. If you put the values you found for $A$ and $B$ into that third equation, you will find it is not identically true. Try a simpler problem, like $\int{x+1\over x}\,dx$, it may be clearer what goes wrong with your approach. – Gerry Myerson Aug 11 '11 at 22:54
• The original rational function goes to infinity as $x$ goes to infinity. The expressions you reduce it to go to $0$, so can't be right. You should always first divide whenever the degree of the "top" is greater than or equal the degree of the "bottom." – André Nicolas Aug 12 '11 at 0:33

Normally one uses the division algorithm to decompose into integral and proper fractional parts

$$\rm \dfrac{2x^3-4x^2-15x+5}{(x+2)\:(x-4)}\ =\ 2\:x + \dfrac{x+5}{(x+2)\:(x-4)}$$

Then one performs the partial fraction decomposition only on the second fractional part. But you seem to desire to skip the initial division algorithm step and, instead, jump immediately to the partial fraction decomposition, proceeding essentially as follows.

$$\rm \dfrac{2x^3-4x^2-15x+5}{(x+2)\:(x-4)}\ = \ \dfrac{A}{x-4}\ +\ \dfrac{B}{x+2}\ +\ \: C(x)$$

$$\rm \Rightarrow\ \ \ 2\:x^3-4\:x^2-15\:x+5\ = \ A\ (x+2) + B\ (x-4)\ +\ C(x)\:(x+2)\:(x-4)$$

Using the Heaviside cover-up method will still work to find the correct values of $\rm\:A\:$ and $\rm\:B\:$ because the term $\rm\:C(x)\:(x+2)\:(x-4)\:$ plays no role - it vanishes when evaluating the RHS at $\rm\:x = 4\:$ and $\rm\:x = -2\:$ while solving for $\rm\:A\:$ and $\rm\:B\:.\:$ Similarly for the LHS which is $\rm\:x+5 + 2\:x\:(x+2)\:(x-4)\:.\:$ Therefore you'll get the same equations for $\rm\:A\:$ and $\rm\:B\:$ whether you use this method or the standard method. Thus, as you do, one can safely set $\rm\:C(x) = 0\:$ from the start. But that yields more work than the standard way, since then you have to evaluate a cubic (vs. $\rm\:x+5\:$) on the LHS. Finally, to obtain the complete answer, you still have to calculate the "integral" part $\rm\:C(x) = 2\:x\:$ and add it to the partial fraction decomposition. The failure to do so explains why your answer is erroneous. You could get the integral part by subtracting the fractional part from the original expression, but that too would be more work than using the division algorithm from the start.

• The "Heaviside cover up" method? – Gerry Myerson Aug 12 '11 at 1:20
• @Gerry See e.g. the Wiki page and see my post here for a nonlinear generalization. – Bill Dubuque Aug 12 '11 at 2:12
• Never knew this method had a name. Live and learn. What did Heaviside have to do with it? Not mentioned in his wikipedia bio, nor the MacTutor. – Gerry Myerson Aug 12 '11 at 4:07