We are tasked with solving the integral:

$\int_1^{e^3} \frac{\ln^4(x^2)}{x} \, dx$

Step-by-step Solution:

1. Simplify the integrand:

   The term $\ln(x^2)$ can be rewritten using logarithmic properties: $\ln(x^2) = 2\ln(x)$ So, $\ln^4(x^2)$ becomes: $\ln^4(x^2) = (2\ln(x))^4 = 16\ln^4(x)$ Thus, the integrand becomes: $\frac{\ln^4(x^2)}{x} = \frac{16\ln^4(x)}{x}$

2. Substitute and rewrite the integral:

   The integral is now: $16 \int_1^{e^3} \frac{\ln^4(x)}{x} \, dx$

3. Use substitution:

   Let $u = \ln(x)$, which implies $du = \frac{dx}{x}$. The limits of integration change as follows:

   • When $x = 1$, $u = \ln(1) = 0$
   • When $x = e^3$, $u = \ln(e^3) = 3$

   The integral becomes: $16 \int_0^3 u^4 \, du$

4. Integrate:

   The integral of $u^4$ is: $\int u^4 \, du = \frac{u^5}{5}$ Applying the limits from 0 to 3: $16 \left[ \frac{u^5}{5} \right]_0^3 = 16 \times \left( \frac{3^5}{5} - \frac{0^5}{5} \right)$ $= 16 \times \frac{243}{5} = \frac{16 \times 243}{5} = \frac{3888}{5}$

Thus, the value of the integral is:

$\boxed{\frac{3888}{5}}$

Would you like further details on any of these steps?

Related Questions:

1. What if the limits of the integral were different?
2. How does logarithmic substitution work in more complex integrals?
3. What are the properties of definite integrals when involving logarithmic functions?
4. How would this problem change if the power of the logarithm were different?
5. Can this method be extended to integrals involving other functions, like exponential terms?

Tip:

When dealing with logarithmic terms in integrals, always look for opportunities to simplify using logarithmic identities like $\ln(x^2) = 2\ln(x)$.