<s>[INST] 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 
 [/INST] 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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