Compute a partial fractions decomposition. 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 
 A: 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
A: 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:
$$\frac{-1}{x}+\frac{2}{x-1} = \frac{x+1}{x(x-1)}$$
A: $x^2-x$ 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
$f(x)=\dfrac{x+1}{x(x-1)}=\dfrac{A}{x}+\dfrac{B}{x-1}$
A: 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 $\;2=B$.

A: \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}
