To find the average value of the function $f(x) = x \ln(x^2) + 8$ on the interval $[e, e^2]$, we use the formula for the average value of a function $f(x)$ over an interval $[a, b]$:

$\text{Average value of } f(x) = \frac{1}{b - a} \int_a^b f(x) \, dx$

Step 1: Simplify the function $f(x)$

The given function is $f(x) = x \ln(x^2) + 8$. Using the logarithmic identity $\ln(x^2) = 2\ln(x)$, we can rewrite the function as:

$f(x) = x \cdot 2\ln(x) + 8 = 2x \ln(x) + 8$

Step 2: Set up the integral

Now, we need to calculate the integral of $f(x)$ from $e$ to $e^2$. This is:

$\int_e^{e^2} \left( 2x \ln(x) + 8 \right) \, dx$

We can split the integral into two parts:

$\int_e^{e^2} 2x \ln(x) \, dx + \int_e^{e^2} 8 \, dx$

Step 3: Compute the first integral $\int_e^{e^2} 2x \ln(x) \, dx$

To solve this, we use integration by parts. Let $u = \ln(x)$ and $dv = 2x \, dx$. Then, $du = \frac{1}{x} \, dx$ and $v = x^2$. Applying the integration by parts formula:

$\int u \, dv = uv - \int v \, du$

We get:

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

Now, $\int x \, dx = \frac{x^2}{2}$, so:

$\int 2x \ln(x) \, dx = x^2 \ln(x) - \frac{x^2}{2}$

Evaluating this from $e$ to $e^2$:

$\left[ x^2 \ln(x) - \frac{x^2}{2} \right]_e^{e^2}$

At $x = e^2$:

$\left( (e^2)^2 \ln(e^2) - \frac{(e^2)^2}{2} \right) = (e^4 \cdot 2\ln(e)) - \frac{e^4}{2} = 0 - \frac{e^4}{2} = -\frac{e^4}{2}$

At $x = e$:

$\left( e^2 \ln(e) - \frac{e^2}{2} \right) = e^2 \cdot 1 - \frac{e^2}{2} = \frac{e^2}{2}$

So, the result of the first integral is:

$-\frac{e^4}{2} - \frac{e^2}{2} = -\frac{e^4 + e^2}{2}$

Step 4: Compute the second integral $\int_e^{e^2} 8 \, dx$

This is straightforward:

$\int_e^{e^2} 8 \, dx = 8 \left[ x \right]_e^{e^2} = 8(e^2 - e) = 8e(e - 1)$

Step 5: Combine the results

The total integral is:

$-\frac{e^4 + e^2}{2} + 8e(e - 1)$

Now, calculate the average value:

$\text{Average value} = \frac{1}{e^2 - e} \left( -\frac{e^4 + e^2}{2} + 8e(e - 1) \right)$

Simplifying this expression will give you the final average value. Would you like me to proceed with further simplification and a numerical result?