Let's solve the integral:

$\int \frac{x^2 + 2x - 25}{(x+2)(x+10)(8x-2)} \, dx$

Step 1: Factor the Denominator

The denominator is already factored as:

$(x+2)(x+10)(8x-2)$

We can simplify $8x - 2$ to $2(4x - 1)$ for ease of working.

So the integral becomes:

$\int \frac{x^2 + 2x - 25}{(x+2)(x+10)2(4x-1)} \, dx$

Step 2: Partial Fraction Decomposition

The next step is to express the integrand as a sum of simpler fractions. We'll decompose the rational function into partial fractions:

$\frac{x^2 + 2x - 25}{(x+2)(x+10)(8x-2)} = \frac{A}{x+2} + \frac{B}{x+10} + \frac{C}{4x-1}$

Step 3: Solve for Constants $A$, $B$, and $C$

Multiply both sides of the equation by the denominator $(x+2)(x+10)(8x-2)$:

$x^2 + 2x - 25 = A(x+10)(8x-2) + B(x+2)(8x-2) + C(x+2)(x+10)$

Now, plug in values for $x$ that make the factors $x+2$, $x+10$, and $4x-1$ zero to solve for $A$, $B$, and $C$.

1. Plugging $x = -2$: $x^2 + 2x - 25 = A(8(-2) - 2)$ $4 - 4 - 25 = A(-18) \Rightarrow A = \frac{-25}{-18} = \frac{25}{18}$

2. Plugging $x = -10$: $x^2 + 2x - 25 = B(4(-10) - 1)$ $100 - 20 - 25 = B(-41) \Rightarrow B = \frac{55}{41}$

3. Plugging $x = \frac{1}{4}$: $x^2 + 2x - 25 = C((\frac{1}{4} + 2)(\frac{1}{4} + 10))$ $(\frac{1}{16} + \frac{2}{4} - 25) = C( \left(\frac{9}{4}\right)(\frac{41}{4}) ) \Rightarrow C=\text{value}$

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