Answer
$\frac{-{{x}^{2}}+4x+1}{\left( x-1 \right)\left( x+3 \right)}$.
Work Step by Step
$\frac{1}{x-1}-\frac{x-2}{x+3}$,
LCD (Least common denominator) of the denominator is
$\text{LCD}=\left( x-1 \right)\left( x+3 \right)$
$\begin{align}
& \frac{1}{x-1}-\frac{x-2}{x+3}=\frac{1}{x-1}\cdot \frac{x+3}{x+3}-\frac{x-2}{x+3}\cdot \frac{x-1}{x-1} \\
& =\frac{x+3}{\left( x-1 \right)\left( x+3 \right)}-\frac{\left( x-2 \right)\left( x-1 \right)}{\left( x+3 \right)\left( x-1 \right)} \\
& =\frac{\left( x+3 \right)-\left( x-1 \right)\left( x-2 \right)}{\left( x-1 \right)\left( x+3 \right)}
\end{align}$
Apply the distributive property and simplify the expression,
$\begin{align}
& \frac{1}{x-1}-\frac{x-2}{x+3}=\frac{\left( x+3 \right)-\left( x\cdot x+x\cdot \left( -2 \right)-1\cdot \left( x \right)-1\cdot \left( -2 \right) \right)}{\left( x-1 \right)\left( x+3 \right)} \\
& =\frac{\left( x+3 \right)-\left( {{x}^{2}}-2x-x+2 \right)}{\left( x-1 \right)\left( x+3 \right)} \\
& =\frac{\left( x+3 \right)-\left( {{x}^{2}}-3x+2 \right)}{\left( x-1 \right)\left( x+3 \right)}
\end{align}$
Apply the rule of subtraction of polynomials,
$\begin{align}
& \frac{1}{x-1}-\frac{x-2}{x+3}=\frac{x+3-{{x}^{2}}+3x-2}{\left( x-1 \right)\left( x+3 \right)} \\
& =\frac{-{{x}^{2}}+4x+1}{\left( x-1 \right)\left( x+3 \right)}
\end{align}$
Therefore, the simplified form of the expression is $\frac{-{{x}^{2}}+4x+1}{\left( x-1 \right)\left( x+3 \right)}$.
