The given expression is:

$\frac{a+1}{\log_{n^3 - n} n} + \frac{a-2}{\log_{n} n}.$

Step-by-step solution:

1. Simplifying the Logarithms:

   • The term $\log_{n^3 - n} n$ can be simplified using the change of base formula for logarithms. The change of base formula states: $\log_{b} a = \frac{\log a}{\log b}.$ Therefore: $\log_{n^3 - n} n = \frac{\log n}{\log(n^3 - n)}.$

2. Simplifying $\log_n n$:

   • The logarithm $\log_{n} n$ is always 1 because: $\log_n n = 1.$

3. Substitute into the Expression:

   • Now substitute the simplifications back into the expression: $\frac{a+1}{\frac{\log n}{\log(n^3 - n)}} + \frac{a-2}{1}.$
   • This can be further simplified by multiplying both numerator and denominator of the first term by $\log(n^3 - n)$: $(a+1) \frac{\log(n^3 - n)}{\log n} + (a-2).$

Final Expression:

The simplified expression becomes: $(a+1) \frac{\log(n^3 - n)}{\log n} + (a-2).$

Would you like to explore further details or have any additional questions on this?

Here are 5 related questions:

1. How do logarithmic properties simplify complex expressions?
2. What is the change of base formula and its applications?
3. Can logarithms of different bases always be converted using common logarithms?
4. What are common logarithmic identities useful in calculus and algebra?
5. How do you handle logarithms involving polynomials or exponential expressions?

Tip: When working with logarithms, remember to check for any special cases (like $\log_b b = 1$) that can simplify your calculations.