https://file.aiquickdraw.com/mathbot/file/1727847116095k5vh422j.jpg

The integral in the image you've provided is:

\[
\int \frac{x^2 - 1}{(x - 2)(x^2 - 4)} \, dx
\]

### Step 1: Factor the Denominator

We can simplify the denominator first. Notice that \(x^2 - 4\) is a difference of squares:
\[
x^2 - 4 = (x - 2)(x + 2)
\]
So, the denominator becomes:
\[
(x - 2)(x - 2)(x + 2)
\]
This simplifies to:
\[
(x - 2)^2(x + 2)
\]

### Step 2: Simplify the Integral

The integral now looks like this:
\[
\int \frac{x^2 - 1}{(x - 2)^2(x + 2)} \, dx
\]
Now, we can break this into partial fractions.

### Step 3: Use Partial Fractions

We decompose the rational function as follows:
\[
\frac{x^2 - 1}{(x - 2)^2(x + 2)} = \frac{A}{x - 2} + \frac{B}{(x - 2)^2} + \frac{C}{x + 2}
\]
To find \(A\), \(B\), and \(C\), multiply both sides by \((x - 2)^2(x + 2)\), which gives:
\[
x^2 - 1 = A(x - 2)(x + 2) + B(x + 2) + C(x - 2)^2
\]
Now, expand the terms on the right-hand side:
\[
A(x^2 - 4) + B(x + 2) + C(x^2 - 4x + 4)
\]
Simplify further:
\[
A(x^2 - 4) + Bx + 2B + C(x^2 - 4x + 4)
= (A + C)x^2 + (B - 4C)x + (-4A + 2B + 4C)
\]
Now, compare coefficients with the left-hand side \(x^2 - 1\):

- For \(x^2\): \(A + C = 1\)
- For \(x\): \(B - 4C = 0\)
- For the constant term: \(-4A + 2B + 4C = -1\)

### Step 4: Solve for \(A\), \(B\), and \(C\)

From \(A + C = 1\), we get \(C = 1 - A\).

Substitute \(C = 1 - A\) into \(B - 4C = 0\):
\[
B - 4(1 - A) = 0 \implies B = 4 - 4A
\]

Now substitute \(B = 4 - 4A\) and \(C = 1 - A\) into \(-4A + 2B + 4C = -1\):
\[
-4A + 2(4 - 4A) + 4(1 - A) = -1
\]
Simplify:
\[
-4A + 8 - 8A + 4 - 4A = -1
\]
\[
-16A + 12 = -1 \implies -16A = -13 \implies A = \frac{13}{16}
\]

Now, find \(B\) and \(C\):
\[
C = 1 - \frac{13}{16} = \frac{3}{16}, \quad B = 4 - 4 \times \frac{13}{16} = \frac{12}{16} = \frac{3}{4}
\]

### Step 5: Substitute Back

So, the partial fraction decomposition is:
\[
\frac{x^2 - 1}{(x - 2)^2(x + 2)} = \frac{13}{16(x - 2)} + \frac{3}{4(x - 2)^2} + \frac{3}{16(x + 2)}
\]

### Step 6: Integrate Each Term

Now, integrate each term separately:

1. \(\int \frac{13}{16(x - 2)} \, dx = \frac{13}{16} \ln|x - 2|\)
2. \(\int \frac{3}{4(x - 2)^2} \, dx = -\frac{3}{4(x - 2)}\)
3. \(\int \frac{3}{16(x + 2)} \, dx = \frac{3}{16} \ln|x + 2|\)

### Step 7: Final Solution

The integral is:
\[
\frac{13}{16} \ln|x - 2| - \frac{3}{4(x - 2)} + \frac{3}{16} \ln|x + 2| + C
\]

Let me know if you'd like any further details or explanations!

### Related Questions:
1. How does partial fraction decomposition work in general?
2. What happens if the denominator has complex roots?
3. How do you integrate terms with powers of linear expressions?
4. How would this problem change if there was a higher power in the denominator?
5. Can you solve similar integrals without using partial fractions?

**Tip:** Always check if you can factor the denominator before attempting partial fractions.

Learn how to solve the integral ∫ (x^2 - 1) / [(x - 2)(x^2 - 4)] dx using partial fraction decomposition. This problem involves factoring the denominator, using partial fractions, and performing logarithmic and rational function integration. Ideal for undergraduate calculus students.

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