To solve the integral:

\[
\int 5 \ln(x^2 - 1) \, dx
\]

we can use **integration by parts**. The formula for integration by parts is:

\[
\int u \, dv = u v - \int v \, du
\]

1. Let \( u = \ln(x^2 - 1) \), so \( du = \frac{2x}{x^2 - 1} \, dx \).
2. Let \( dv = 5 \, dx \), so \( v = 5x \).

Now, applying the formula:

\[
\int 5 \ln(x^2 - 1) \, dx = 5x \ln(x^2 - 1) - \int 5x \cdot \frac{2x}{x^2 - 1} \, dx
\]

Simplifying the second integral:

\[
\int \frac{10x^2}{x^2 - 1} \, dx
\]

Next, we split the fraction:

\[
\int \left( 10 + \frac{10}{x^2 - 1} \right) \, dx = 10x + 10 \int \frac{1}{x^2 - 1} \, dx
\]

The integral of \( \frac{1}{x^2 - 1} \) can be expressed as a difference of logarithms:

\[
\int \frac{1}{x^2 - 1} \, dx = \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right|
\]

Now we combine everything:

\[
\int 5 \ln(x^2 - 1) \, dx = 5x \ln(x^2 - 1) - 10x - 5 \ln \left| \frac{x - 1}{x + 1} \right| + C
\]

where \( C \) is the constant of integration.

---

Would you like more details, or do you have any questions?

Here are 5 related questions to explore:

1. How is integration by parts applied to logarithmic functions in other cases?
2. Can you use substitution to solve integrals like this?
3. How would the integral change if the argument of the logarithm was different, say \( \ln(x^2 + 1) \)?
4. What are the applications of logarithmic integrals in physics or engineering?
5. How can partial fraction decomposition help in solving rational integrals?

**Tip:** When facing complex logarithmic integrals, remember that splitting the logarithmic expression into simpler terms can often simplify the integration process.

Explore how to solve the integral of 5 ln(x^2 − 1) dx using integration by parts. This problem involves key calculus concepts such as logarithmic functions, partial fractions, and integration by parts. Suitable for undergraduate students studying Calculus II.

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