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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.

Evaluate the integral: ∫ from 0 to 1 of x^3 [log(1/x)]^4 dx

Learn how to evaluate the integral ∫ from 0 to 1 of x^3 [log(1/x)]^4 dx using substitution and known formulas. This problem covers key calculus concepts like logarithmic functions and integration techniques, suitable for university-level students.

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