The given integral is:

$\int_0^1 x^3 \left( \log \left( \frac{1}{x} \right) \right)^4 \, dx$

Steps to solve:

1. Simplify the logarithmic term: $\log \left( \frac{1}{x} \right) = -\log(x)$ Thus, the integral becomes: $\int_0^1 x^3 \left( -\log(x) \right)^4 \, dx$

   Since $(- \log(x))^4 = \log(x)^4$, we have: $\int_0^1 x^3 \log(x)^4 \, dx$

2. Substitute: Let $u = \log(x)$, so $du = \frac{1}{x} dx$ and $x = e^u$.

   • When $x = 0$, $u = -\infty$
   • When $x = 1$, $u = 0$

   Also, $x^3 = e^{3u}$. Hence, the integral becomes: $\int_{-\infty}^0 e^{3u} u^4 \, du$

3. Solving the integral: This integral can be solved using integration by parts or by using a known formula for integrals of the form $\int e^{au} u^n \, du$. The general result for such integrals is:

   $\int e^{au} u^n \, du = \frac{n!}{a^{n+1}}$

   In this case, $a = 3$ and $n = 4$. Applying the formula:

   $\int_{-\infty}^0 e^{3u} u^4 \, du = \frac{4!}{3^5} = \frac{24}{243} = \frac{8}{81}$

Thus, the value of the integral is:

$\boxed{\frac{8}{81}}$

Would you like any further clarification on the steps or a deeper explanation? Here are 5 related questions to expand your understanding:

1. How does the substitution $u = \log(x)$ affect the bounds of integration?
2. Can you generalize the result for $\int_0^1 x^n (\log(1/x))^k \, dx$?
3. What role does integration by parts play in solving more complex integrals involving $\log(x)$?
4. How does the factorial term arise in the integral of $e^{au} u^n$?
5. Can this technique be extended to solve integrals involving higher powers of logarithmic functions?

Tip: When dealing with logarithmic terms in integrals, substitution is often a helpful first step to simplify the expression.