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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

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

  • 0
    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.2011-08-11
  • 0
    You should do long division first.2011-08-11
  • 1
    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.2011-08-11
  • 0
    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."2011-08-12

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