To compute the indefinite integral of:

$\int \left( x^2 + \frac{1}{(x-9)(x-8)^2} \right) dx$

We can break it down into two parts:

1. The integral of $x^2$.
2. The integral of $\frac{1}{(x-9)(x-8)^2}$.

Step 1: Integrating $x^2$

$\int x^2 dx = \frac{x^3}{3}$

Step 2: Integrating $\frac{1}{(x-9)(x-8)^2}$

To handle this, we apply partial fraction decomposition. We express $\frac{1}{(x-9)(x-8)^2}$ as:

$\frac{1}{(x-9)(x-8)^2} = \frac{A}{x-9} + \frac{B}{x-8} + \frac{C}{(x-8)^2}$

Now, let's solve for $A$, $B$, and $C$.

$\frac{1}{(x-9)(x-8)^2} = \frac{A(x-8)^2 + B(x-9)(x-8) + C(x-9)}{(x-9)(x-8)^2}$

This simplifies to:

$1 = A(x-8)^2 + B(x-9)(x-8) + C(x-9)$

Now, we expand and solve for $A$, $B$, and $C$.

Let me compute that.The partial fraction decomposition gives:

$\frac{1}{(x-9)(x-8)^2} = \frac{1}{x-9} - \frac{1}{x-8} - \frac{1}{(x-8)^2}$

Step 3: Integrating the decomposition

Now, we integrate each term:

1. $\int \frac{1}{x-9} dx = \ln|x-9|$
2. $\int \frac{1}{x-8} dx = \ln|x-8|$
3. $\int \frac{1}{(x-8)^2} dx = -\frac{1}{x-8}$

Step 4: Final Answer

The full indefinite integral is:

$\int \left( x^2 + \frac{1}{(x-9)(x-8)^2} \right) dx = \frac{x^3}{3} + \ln|x-9| - \ln|x-8| + \frac{1}{x-8} + C$

where $C$ is the constant of integration.

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