In Problems 11–20, determine the partial fraction expansion for the given rational function.

$\frac{\mathbf{-}\mathbf{s}\mathbf{-}\mathbf{7}}{\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{2}\mathbf{)}}$

• The partial fraction expansion of a rational fraction (that is, a fraction with both a polynomial numerator and a polynomial denominator) is an algebraic procedure that involves expressing the fraction as a sum of a polynomial plus one or more fractions with a simpler denominator.
• The partial fraction decomposition is important because it gives methods for a variety of rational function computations, such as explicit antiderivative computations, Taylor series expansions, inverse Z-transforms, and inverse Laplace transforms.

The given rational function is $\frac{-\mathrm{s}-7}{(\mathrm{s}+1)(\mathrm{s}-2)}$

Rewrite $\frac{-\mathrm{s}-7}{(\mathrm{s}+1)(\mathrm{s}-2)}$ as a sum of partial fractions as:

$\frac{-\mathrm{s}-7}{(\mathrm{s}+1)(\mathrm{s}-2)}=\frac{\mathrm{A}}{\mathrm{s}+1}+\frac{\mathrm{B}}{\mathrm{s}-2}$

Multiply both sides by the LCD $(\mathrm{s}+1)(\mathrm{s}-2)$as follows:

$-\mathrm{s}-7=\mathrm{A}(\mathrm{s}-2)+\mathrm{B}(\mathrm{s}+1)$

Find the constants as:

For$\mathrm{s}=-1,-(-1)-7=\mathrm{A}(-3)\Rightarrow \mathrm{A}=2$.

For $\mathrm{s}=2,-\left(2\right)-7=\mathrm{B}\left(3\right)\Rightarrow \mathrm{B}=-3$.

Substitute the value of constants into $\frac{-\mathrm{s}-7}{(\mathrm{s}+1)(\mathrm{s}-2)}=\frac{\mathrm{A}}{\mathrm{s}+1}+\frac{\mathrm{B}}{\mathrm{s}-2}$as follows:

$\begin{array}{c}\frac{-\mathrm{s}-7}{(\mathrm{s}+1)(\mathrm{s}-2)}=\frac{2}{\mathrm{s}+1}+\frac{-3}{\mathrm{s}-2}\\ =\frac{2}{\mathrm{s}+1}-\frac{3}{\mathrm{s}-2}\end{array}$

Therefore, the partial fraction expansion for the given rational function is $\frac{2}{\mathrm{s}+1}-\frac{3}{\mathrm{s}-2}$.