The integral shown is:

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

Approach:

We will solve this integral using partial fraction decomposition. Here are the steps:

1. Partial Fraction Decomposition: The goal is to express the integrand as a sum of simpler fractions. Since the denominator is a product of three linear terms $(x+2)$, $(x+10)$, and $(8x-2)$, we assume the form of the partial fraction decomposition as:

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

2. Solving for $A$, $B$, and $C$: Multiply both sides by the denominator $(x+2)(x+10)(8x-2)$ to eliminate the denominator:

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

   Now expand both sides, compare coefficients, and solve for $A$, $B$, and $C$.

3. Integrate each term: After finding $A$, $B$, and $C$, integrate each partial fraction individually, which will result in logarithmic terms.

Let me calculate the partial fraction decomposition for this integral.The partial fraction decomposition yields the following coefficients:

$A = \frac{25}{144}, \quad B = \frac{55}{656}, \quad C = -\frac{391}{369}$

Thus, the integrand can be rewritten as:

$\frac{x^2 + 2x - 25}{(x+2)(x+10)(8x-2)} = \frac{25}{144(x+2)} + \frac{55}{656(x+10)} - \frac{391}{369(8x-2)}$

Step 2: Integrate each term

Now, the integral becomes:

$\int \frac{25}{144(x+2)} \, dx + \int \frac{55}{656(x+10)} \, dx - \int \frac{391}{369(8x-2)} \, dx$

I will now compute these integrals.The integrals of each term are:

$\int \frac{25}{144(x+2)} \, dx = \frac{25}{144} \ln(x+2)$

$\int \frac{55}{656(x+10)} \, dx = \frac{55}{656} \ln(x+10)$

$\int \frac{-391}{369(8x-2)} \, dx = \frac{-391}{2952} \ln(8x-2)$

Final Solution:

The complete integral is:

$\frac{25}{144} \ln(x+2) + \frac{55}{656} \ln(x+10) - \frac{391}{2952} \ln(8x-2) + C$

Where $C$ is the constant of integration.

Would you like any further details or clarifications?

Related Questions:

1. How would you solve an integral involving a higher-degree polynomial in the denominator?
2. Can you explain how partial fraction decomposition works for repeated factors?
3. How does logarithmic integration arise in these types of problems?
4. What happens if the numerator has a higher degree than the denominator in an integral?
5. Could you apply this technique to integrals involving trigonometric functions?

Tip:

When integrating with logarithms, always check that the argument inside the logarithm is positive to ensure the result is valid within the domain of the function.