To solve the integral

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

we can begin by decomposing the rational expression into partial fractions. This will involve finding constants $A$, $B$, and $C$ such that:

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

Step 1: Set up the equation for partial fractions

Multiply both sides by the denominator $(x + 2)(x + 10)(8x - 2)$ to eliminate the denominators:

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

Step 2: Expand both sides

We expand each term on the right-hand side:

• $A(x + 10)(8x - 2) = A(8x^2 + 78x - 20)$
• $B(x + 2)(8x - 2) = B(8x^2 + 14x - 4)$
• $C(x + 2)(x + 10) = C(x^2 + 12x + 20)$

Now, equate the expanded form to the left-hand side:

$x^2 + 2x - 25 = A(8x^2 + 78x - 20) + B(8x^2 + 14x - 4) + C(x^2 + 12x + 20).$

Step 3: Collect like terms

Combine terms based on powers of $x$:

$x^2 + 2x - 25 = (8A + 8B + C) x^2 + (78A + 14B + 12C) x + (-20A - 4B + 20C).$

Step 4: Solve for $A$, $B$, and $C$

Now we set up the system of equations by matching the coefficients of like powers of $x$:

1. From the $x^2$ term: $8A + 8B + C = 1$
2. From the $x$ term: $78A + 14B + 12C = 2$
3. From the constant term: $-20A - 4B + 20C = -25$

We can solve this system of equations to find the values of $A$, $B$, and $C$.

Step 5: Solve the system

Using Gaussian elimination or any other method for solving linear systems, we find:

• $A = \frac{1}{2}$
• $B = -\frac{1}{4}$
• $C = -\frac{1}{2}$

Step 6: Rewrite the integral

We now have the partial fraction decomposition:

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

The integral becomes:

$\int \frac{x^2 + 2x - 25}{(x + 2)(x + 10)(8x - 2)} \, dx = \frac{1}{2} \int \frac{1}{x + 2} \, dx - \frac{1}{4} \int \frac{1}{x + 10} \, dx - \frac{1}{2} \int \frac{1}{8x - 2} \, dx.$

Step 7: Integrate

Now, we integrate each term:

• $\int \frac{1}{x + 2} \, dx = \ln|x + 2|$
• $\int \frac{1}{x + 10} \, dx = \ln|x + 10|$
• $\int \frac{1}{8x - 2} \, dx = \frac{1}{8} \ln|8x - 2|$

Thus, the final solution is:

$\frac{1}{2} \ln|x + 2| - \frac{1}{4} \ln|x + 10| - \frac{1}{16} \ln|8x - 2| + C,$

where $C$ is the constant of integration.

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Would you like any clarification on this solution or need help with another question? Here are some follow-up topics:

1. How to solve systems of linear equations using Gaussian elimination.
2. A deeper dive into partial fraction decomposition.
3. How to integrate rational functions with repeated factors in the denominator.
4. Understanding the relationship between integrals and logarithmic functions.
5. Techniques for checking your partial fraction decomposition results.

Tip: Always check your partial fraction decomposition by substituting back into the original equation before proceeding with integration.