To express the given rational function

$\frac{x+5}{(x-1)(x+2)}$

as partial fractions, we follow these steps:

Step 1: Decompose into Partial Fractions

We assume the decomposition of the form:

$\frac{x+5}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$

Here, $A$ and $B$ are constants we need to determine.

Step 2: Clear the Denominator

Multiply both sides of the equation by $(x-1)(x+2)$ to eliminate the denominators:

$x + 5 = A(x + 2) + B(x - 1)$

Step 3: Expand and Simplify

Now expand both sides:

$x + 5 = A(x + 2) + B(x - 1)$

$x + 5 = A(x) + 2A + B(x) - B$

$x + 5 = (A + B)x + (2A - B)$

Step 4: Equate Coefficients

Now, equate the coefficients of like powers of $x$ on both sides. The equation is:

$x + 5 = (A + B)x + (2A - B)$

For the $x$-terms:
$1 = A + B$

For the constant terms:
$5 = 2A - B$

Step 5: Solve the System of Equations

We now have a system of two equations:

1. $A + B = 1$
2. $2A - B = 5$

Solve this system:

From $A + B = 1$, we get $B = 1 - A$. Substitute this into the second equation:

$2A - (1 - A) = 5$

$2A - 1 + A = 5$

$3A = 6$

$A = 2$

Now substitute $A = 2$ into $B = 1 - A$:

$B = 1 - 2 = -1$

Step 6: Final Partial Fraction Decomposition

Thus, the partial fraction decomposition of $\frac{x+5}{(x-1)(x+2)}$ is:

$\frac{x+5}{(x-1)(x+2)} = \frac{2}{x-1} - \frac{1}{x+2}$

Conclusion:

$\frac{x+5}{(x-1)(x+2)} = \frac{2}{x-1} - \frac{1}{x+2}$

Would you like further details on any part of the solution or have any questions?

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Here are 5 follow-up questions to explore further:

1. How do you handle partial fractions when the denominator has repeated factors?
2. What is the process if the denominator is a higher-degree polynomial?
3. How would you use partial fraction decomposition in solving integrals?
4. Can partial fraction decomposition be applied to irreducible quadratic factors?
5. How do you solve for constants when the right-hand side includes both linear and quadratic terms?

Tip: When solving for partial fractions, always check that the degree of the numerator is less than the degree of the denominator before proceeding!