Integration by partial fraction mistake

I am trying to understand my mistake using the partial fraction method to solve $\displaystyle\int{\frac{2x-5}{x^2-4x+4}}$.

Here's what I have so far: $$\frac{2x+5}{x^3-4x^2+4x} = \frac{2x+5}{x(x-2)^2} = \frac{A}{x} + \frac{B}{(x-2)^2}$$

And its easy to see that $A=\displaystyle\frac{5}{4}$ and that $B=\displaystyle\frac{9}{2}$. Somehow I get a completely different result then that of my professor's. She did the following:

$$\frac{2x+5}{x^3-4x^2+4x} = \frac{2x+5}{x(x-2)^2} = \frac{A}{x} + \frac{B}{(x-2)} + \frac{C}{(x-2)^2}$$

Can anyone please tell me where my mistake is? Thanks

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There must be a mistake in your notation, because $x^2-4x+4$ is not $x(x-2)^2$. – Potato Jan 7 '12 at 11:14
$x^2-4x +4 \neq x(x-2)^2$ – pedja Jan 7 '12 at 11:14
It was a typing mistake, thanks. The question remains! – yotamoo Jan 7 '12 at 11:18
partial fraction – pedja Jan 7 '12 at 11:21
what is the right denominator $x^3-4x^2+4x$ or $x^2-4x+4$ – iostream007 May 10 '13 at 7:20

You might note that your decomposition does not work: $$\tag{1}\frac{2x+5}{x(x-2)^2} \ne \frac{5/4}{\vphantom{)^2}x} + \frac{9/2}{(x-2)^2};$$ because, while there is no $x^2$ term in the numerator of the left hand side of (1), when you do the addition on the right hand side, its numerator will contain a non-zero $x^2$-term.

In fact, no decomposition of the form $$\tag{2}\frac{2x+5}{x(x-2)^2} = \frac{A}{\vphantom{)^2}x} + \frac{B}{(x-2)^2}$$ would work. If you perform the addition on the right, it is clear that $A$ must be 0 (otherwise there would be a non-zero $x^2$ term on the right); but then no value of $B$ would give the correct numerator.

You may be lead to surmise that, in order for the method at hand to succeed, one more additive rational term is needed in the right hand side of (2). The denominator of this term should be different from the others on the right hand side of (2) and "part of" the denominator on the left hand side of (2).

This leads one to consider the decomposition

$$\frac{2x+5}{x(x-2)^2} = \frac{A}{x\vphantom{)^2}}+ \frac{B}{\vphantom{)^2} x-2 } + \frac{C}{(x-2)^2};$$ which is, indeed, the correct decomposition (in particular, this decomposition allows you to obtain the correct $x^2$ term on the left (0, here) ).

The above is not meant to be rigorous, but was an attempt to illustrate why the decomposition above should contain the ${B\over x-2}$ term.

More generally, when writing the form of a partial fraction decomposition, you will want to express the given rational expression as a sum of "simpler" rational expressions in the most general way possible. The rule below (whose proof I'll leave to those with more expertise than myself) makes explicit what we mean by "the most general way possible".

Partial Fraction Decomposition

Suppose $P(x)\over Q(x)$ is a rational expression in lowest terms with ${\rm deg}\, P\!\lt\!{\rm deg}\, Q$.

To obtain the partial fraction decomposition ${P(x)\over Q(x)}=\Phi$, first set $\Phi=0$; then

• For each distinct factor $(x-a)^{ n}$ of $Q$ where $n$ is maximal, add the term $$\def\ss{}\def\ts{}\def\sss{} { {{\ss A_{\sss1}}\over {\ss x-a\vphantom{)^2}}} +\ts{{\ss A_{\sss2}}\over {\ss(x-a)^{\sss2}}} +\cdots +\ts{{\ss A_{\sss n}}\over {\ss(x-a)^{\sss n\vphantom{2}}}}}$$ to $\Phi$ (of course, use different constants for different factors).
• For each distinct irreducible factor $(x^{\sss2}+bx+c)^{\sss n}$ of $Q$ where $n$ is maximal, add the term $$\def\sss{} {B_{\sss1}x+C_{\sss1}\over \vphantom{)^2} x^{\sss2}+bx+c } +{B_{\sss2}x+C_{\sss2}\over (x^{\sss2}+bx+c)^{\sss2}}+\cdots +{B_{\sss n}x+C_{\sss n}\over \vphantom{)^2} (x^{\sss2}+bx+c)^{\sss n}}$$ to $\Phi$ (as above, use different constants for different factors).
• To find the constants: Write down the PFD, multiply both sides by $Q$ and simplify. Then expand both sides and equate coefficients of powers of $x$. Solve for the constants.

Please see Paul's post for examples.

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If the denominator has no repeated factor, then it's easy to express it into partial fraction. For example $$\frac{-2x+3}{x^3-x} =\frac{-2x+3}{x(x-1)(x+1)}= \frac{A}{x} + \frac{B}{(x-1)} + \frac{C}{(x+1)},$$ $$\frac{7}{x^4-3x^2+2} =\frac{-2x+3}{(x-2)(x+2)(x-1)(x+1)}= \frac{A}{x-2} + \frac{B}{x+2} + \frac{C}{x-1}+\frac{C}{x+1} .$$ However, if the denominator has repeated factor, it's more complicated. Basically we include all the lower power of the repeated factor. Therefore, in your example, we have $$\frac{2x+5}{x^3-4x^2+4x} = \frac{2x+5}{x(x-2)^2} = \frac{A}{x} + \frac{B}{(x-2)} + \frac{C}{(x-2)^2}.$$ Let me do two more examples $$\frac{4x+1}{x^4+2x^3+x^2}=\frac{4x+1}{x^2(x+1)^2}=\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} + \frac{D}{(x+1)^2},$$ $$\frac{-5x+1}{(x+2)(x-1)^3}=\frac{A}{x+2} + \frac{B}{x-1} + \frac{C}{(x-1)^2} + \frac{D}{(x-1)^3}.$$

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