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How do I compute a partial fractions decomposition of $$\dfrac{x+1}{x^2-x}$$

I've tried,

Since, $x^2-x$ is a quadratic expression so, I wrote the partial fraction decomposition of this as $$\dfrac{Ax+B}{x^2-x}$$

But the answer came incorrect. Thanks

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  • $\begingroup$ Please see my ansewr for a full soln $\endgroup$ – K Split X Feb 28 '17 at 13:07
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Partial Fraction Decomposition asks

$$\dfrac{x+1}{x(x-1)}=\dfrac Ax+\dfrac B{x-1}$$

$$x+1=A(x-1)+Bx$$

Put $x=0$ and $1$ one by one

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One of the biggest mistakes people make is not factoring the deonimator completly! You have made the same mistake.

Notice how:$$\dfrac{x+1}{x^2-x} = \dfrac{x+1}{x(x-1)}$$

Now node the $x$ and $x-1$. Both are linear. Therefore the simplest form is juts a constant, lets call it $A$, and $B$.

So you have:

$$\frac{A}{x}+\frac{B}{x-1} = \frac{x+1}{x(x-1)}$$

Now let's solve.

$$\frac{A(x-1)+Bx}{x(x-1)} = \frac{x+1}{x(x-1)}$$

$$\frac{Ax-A+Bx}{x(x-1)} = \frac{x+1}{x(x-1)}$$ $$\frac{(A-B)x-A}{x(x-1)} = \frac{x+1}{x(x-1)}$$

Now equate the equation. From the numerator $x+1$, the degree $1$ term is just the coefficient on $x$, which is $1$, and the degree $0$ term is $+1$. So we have:

$$A-B=1$$ $$-A=1$$

Therefore, $A=-1$, B=2$

Thus, the $p.f.d$ is:

$$\frac{-1}{x}+\frac{2}{x-1} = \frac{x+1}{x(x-1)}$$

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$x^2-x$ can be written as $$x(x-1)$$ $f(x)=\dfrac{x+1}{x(x-1)}$

When $Q(x)$ have distinct linear factors

$\dfrac{P(x)}{Q(x)}=\dfrac{A_1}{a_1x+b_1}+\dfrac{A_2}{a_2x+b_2}+....+\dfrac{A_n}{a_nx+b_n}$

In this case

$f(x)=\dfrac{x+1}{x(x-1)}=\dfrac{A}{x}+\dfrac{B}{x-1}$

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You need a decomposition of the denominator into irreducible factors first:

The partial fraction decomposition of $\dfrac{N(x)}{\underbrace{P_1(x)\dotsm P_r(x)}_{D(x)}}$, where $P_1,\dots,P_r$ are distinct irreducible polynomials and $\deg N(x)<\deg D(x)$ has the form $$\frac{N_1(x)}{P_1(x)}+\dots+\frac{N_r(x)}{P_r(x)}\quad\text{where}\quad \deg(Ni<\deg P_i.$$

Hence in the present case, the irreducible factors have degree $1$: $\;x^2-x=x(x-1)$, the decomposition has the form $$\frac{x+1}{x(x-1)}=\frac Ax+\frac B{x-1}, \quad A, B\in\mathbf R.$$

Determination of $A$ and $B$

Multiplying both sides by $x(x-1)$ to remove the denominators, this equation can be written as $$x+1=A(x-1)+Bx.$$ Now set successively:

  • $x=0$, you get $\;1= -A$, i.e. $\;A=-1$,
  • $x=1$, yielding $\;2=B$.
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\begin{gathered} \frac{{x + 1}}{{{x^2} - x}} = \frac{{x + 1}}{{x \cdot (x - 1)}} \ = \frac{{2 \cdot x + 1 - x}}{{x \cdot (x - 1)}} \ = \frac{{2 \cdot x}}{{x \cdot (x - 1)}} - \frac{{x - 1}}{{x \cdot (x - 1)}} \ = \frac{2}{{x - 1}} - \frac{1}{x} \\ \end{gathered}

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