To solve the integral 

\[
\int \frac{2x^2 + 2x - 18}{x^3 - 6x^2 + 9x} \, dx,
\]

we can simplify the integrand before performing the integration. Let's proceed step by step:

### Step 1: Factor the denominator
We can factor the denominator $x^3 - 6x^2 + 9x$ as:

\[
x^3 - 6x^2 + 9x = x(x^2 - 6x + 9) = x(x - 3)^2.
\]

So the integral becomes:

\[
\int \frac{2x^2 + 2x - 18}{x(x - 3)^2} \, dx.
\]

### Step 2: Use partial fraction decomposition
We now decompose the fraction:

\[
\frac{2x^2 + 2x - 18}{x(x - 3)^2}.
\]

Assume that:

\[
\frac{2x^2 + 2x - 18}{x(x - 3)^2} = \frac{A}{x} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2}.
\]

Multiply both sides by $x(x - 3)^2$ to get:

\[
2x^2 + 2x - 18 = A(x - 3)^2 + Bx(x - 3) + Cx.
\]

Now, expand the right-hand side:

\[
A(x - 3)^2 = A(x^2 - 6x + 9),
\]
\[
Bx(x - 3) = B(x^2 - 3x),
\]
\[
Cx = Cx.
\]

So, combining terms:

\[
2x^2 + 2x - 18 = A(x^2 - 6x + 9) + B(x^2 - 3x) + Cx.
\]

\[
2x^2 + 2x - 18 = (A + B)x^2 + (-6A - 3B + C)x + 9A.
\]

### Step 3: Set up a system of equations
Now, compare the coefficients of powers of $x$ on both sides:

- Coefficient of $x^2$: $A + B = 2$,
- Coefficient of $x$: $-6A - 3B + C = 2$,
- Constant term: $9A = -18$.

From the constant term equation, we have:

\[
9A = -18 \implies A = -2.
\]

Substitute $A = -2$ into the first equation:

\[
-2 + B = 2 \implies B = 4.
\]

Substitute $A = -2$ and $B = 4$ into the second equation:

\[
-6(-2) - 3(4) + C = 2 \implies 12 - 12 + C = 2 \implies C = 2.
\]

### Step 4: Rewrite the integral
We now have the partial fractions:

\[
\frac{2x^2 + 2x - 18}{x(x - 3)^2} = \frac{-2}{x} + \frac{4}{x - 3} + \frac{2}{(x - 3)^2}.
\]

So, the integral becomes:

\[
\int \left( \frac{-2}{x} + \frac{4}{x - 3} + \frac{2}{(x - 3)^2} \right) \, dx.
\]

### Step 5: Integrate each term
Now integrate term by term:

1. $\int \frac{-2}{x} \, dx = -2 \ln|x|$,
2. $\int \frac{4}{x - 3} \, dx = 4 \ln|x - 3|$,
3. $\int \frac{2}{(x - 3)^2} \, dx = \frac{-2}{x - 3}$.

### Step 6: Final answer
Combining all the integrals, we get the final solution:

\[
\int \frac{2x^2 + 2x - 18}{x(x - 3)^2} \, dx = -2 \ln|x| + 4 \ln|x - 3| - \frac{2}{x - 3} + C,
\]

where $C$ is the constant of integration.

Would you like further details or have any questions?

---

Here are 5 related questions:
1. How do we approach partial fraction decomposition in general cases?
2. What are other methods to simplify integrals involving rational functions?
3. Can you explain why logarithmic functions appear in the solution?
4. How do we handle improper integrals involving similar rational functions?
5. How does the factorization of polynomials help in solving integrals?

**Tip:** Always check if you can simplify or factor the integrand before attempting more advanced techniques.

$$ \int \frac{2x^2+2x-18}{x^3-6x^2+9x} $$

Learn how to solve the integral of (2x^2 + 2x - 18)/(x^3 - 6x^2 + 9x) using polynomial factorization and partial fraction decomposition. This detailed solution covers key calculus concepts suitable for students in Grades 11-12 or university-level Calculus I.

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