To solve the integral \( I = \int \frac{2}{x} \cdot x^{\ln x} \cdot (\ln x)^3 \, dx \) and find the ratio \( \frac{A}{B} \) where \( I = A x^{(\ln x)^2} - B x^{\ln x} + C \), we will proceed step by step.
### Step 1: Change of Variables
Let \( y = \ln x \). Then, we have:
\[
x = e^y \quad \text{and} \quad dx = e^y \, dy.
\]
Substituting these into the integral gives:
\[
I = \int \frac{2}{e^y} \cdot (e^y)^{y} \cdot y^3 \cdot e^y \, dy = \int 2 e^{y^2} y^3 \, dy.
\]
### Step 2: Simplifying the Integral
Now, we can rewrite the integral as:
\[
I = 2 \int y^3 e^{y^2} \, dy.
\]
### Step 3: Another Change of Variables
Let \( t = y^2 \). Then, \( dt = 2y \, dy \) or \( dy = \frac{dt}{2\sqrt{t}} \). The integral becomes:
\[
I = 2 \int y^3 e^{y^2} \, dy = 2 \int (t^{3/2}) e^t \cdot \frac{dt}{2\sqrt{t}} = \int t e^t \, dt.
\]
### Step 4: Integration by Parts
Now we will use integration by parts where:
- Let \( u = t \) and \( dv = e^t \, dt \).
- Then, \( du = dt \) and \( v = e^t \).
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[
\int t e^t \, dt = t e^t - \int e^t \, dt = t e^t - e^t + C.
\]
### Step 5: Back Substituting
Now substituting back \( t = y^2 \) and \( y = \ln x \):
\[
I = y^2 e^{y^2} - e^{y^2} + C = (\ln x)^2 x^{\ln x} - x^{\ln x} + C.
\]
### Step 6: Comparing Coefficients
From the expression \( I = A x^{(\ln x)^2} - B x^{\ln x} + C \), we can identify:
- \( A = 1 \)
- \( B = 1 \)
### Step 7: Finding \( \frac{A}{B} \)
Thus, we have:
\[
\frac{A}{B} = \frac{1}{1} = 1.
\]
### Final Answer
The value of \( \frac{A}{B} \) is \( 1 \).
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