To evaluate the integral

$\int \frac{-45x^2 - 27x + 18}{x^3 + 2x^2} \, dx,$

we'll break the process down into steps.

Step 1: Factor the denominator

The denominator $x^3 + 2x^2$ can be factored as:

$x^2(x + 2).$

Thus, the integral becomes:

$\int \frac{-45x^2 - 27x + 18}{x^2(x + 2)} \, dx.$

Step 2: Partial fraction decomposition

We aim to decompose the fraction:

$\frac{-45x^2 - 27x + 18}{x^2(x + 2)}$

into simpler fractions of the form:

$\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 2}.$

Multiplying both sides by the denominator $x^2(x + 2)$, we get:

$-45x^2 - 27x + 18 = A x(x + 2) + B(x + 2) + Cx^2.$

Now, expand the right-hand side:

$A x(x + 2) = A x^2 + 2A x,$ $B(x + 2) = B x + 2B,$ $C x^2 = C x^2.$

Combining all terms:

$A x^2 + 2A x + B x + 2B + C x^2.$

Group terms by powers of $x$:

$(A + C)x^2 + (2A + B)x + 2B.$

Step 3: Equate coefficients

Now, equate the coefficients of powers of $x$ from both sides:

1. For $x^2$: $A + C = -45$,
2. For $x^1$: $2A + B = -27$,
3. For $x^0$: $2B = 18$.

Step 4: Solve the system of equations

From equation (3), solve for $B$:

$2B = 18 \implies B = 9.$

Substitute $B = 9$ into equation (2):

$2A + 9 = -27 \implies 2A = -36 \implies A = -18.$

Substitute $A = -18$ into equation (1):

$-18 + C = -45 \implies C = -27.$

Step 5: Rewrite the integral

Now we can rewrite the fraction as:

$\frac{-45x^2 - 27x + 18}{x^2(x + 2)} = \frac{-18}{x} + \frac{9}{x^2} + \frac{-27}{x + 2}.$

Thus, the integral becomes:

$\int \left( \frac{-18}{x} + \frac{9}{x^2} + \frac{-27}{x + 2} \right) dx.$

Step 6: Integrate term by term

Now integrate each term separately:

1. $\int \frac{-18}{x} \, dx = -18 \ln |x|,$
2. $\int \frac{9}{x^2} \, dx = -\frac{9}{x},$
3. $\int \frac{-27}{x + 2} \, dx = -27 \ln |x + 2|.$

Step 7: Final solution

Thus, the final solution to the integral is:

$-18 \ln |x| - \frac{9}{x} - 27 \ln |x + 2| + C,$

where $C$ is the constant of integration.

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Do you want more details or have any questions?
Here are 5 questions you can explore next:

1. How does partial fraction decomposition work for higher-degree polynomials?
2. Can you try integrating a similar rational function without factorable denominators?
3. How do we know that partial fractions is a good method for this integral?
4. What is the process for factoring higher-degree polynomials?
5. Can partial fractions be applied to improper fractions?

Tip: Always try factoring the denominator first in rational integrals; this often simplifies the problem greatly!