`int_(1//e)^(e) (dx)/(x(log x)^(1//3))`

Answer

To solve the integral

\[ I = \int_{\frac{1}{e}}^{e} \frac{dx}{x (\log x)^{\frac{1}{3}}} \]

we will use the substitution \( \log x = t \).

### Step 1: Change of Variables
Let \( t = \log x \). Then, we differentiate both sides:

\[ \frac{dt}{dx} = \frac{1}{x} \implies dx = x \, dt = e^t \, dt \]

### Step 2: Change the Limits of Integration
When \( x = \frac{1}{e} \),

\[ t = \log\left(\frac{1}{e}\right) = -1 \]

When \( x = e \),

\[ t = \log(e) = 1 \]

So, the limits change from \( x: \left[\frac{1}{e}, e\right] \) to \( t: [-1, 1] \).

### Step 3: Substitute in the Integral
Now substituting \( x = e^t \) into the integral, we have:

\[ I = \int_{-1}^{1} \frac{e^t \, dt}{e^t (t)^{\frac{1}{3}}} = \int_{-1}^{1} \frac{dt}{t^{\frac{1}{3}}} \]

### Step 4: Simplify the Integral
The integral simplifies to:

\[ I = \int_{-1}^{1} t^{-\frac{1}{3}} \, dt \]

### Step 5: Evaluate the Integral
Now we can evaluate the integral:

\[ I = \int_{-1}^{1} t^{-\frac{1}{3}} \, dt \]

This can be split into two parts:

\[ I = \int_{-1}^{0} t^{-\frac{1}{3}} \, dt + \int_{0}^{1} t^{-\frac{1}{3}} \, dt \]

### Step 6: Calculate Each Integral
For \( t^{-\frac{1}{3}} \):

\[ \int t^{-\frac{1}{3}} \, dt = \frac{t^{\frac{2}{3}}}{\frac{2}{3}} = \frac{3}{2} t^{\frac{2}{3}} \]

Evaluating from \( -1 \) to \( 0 \):

\[ \int_{-1}^{0} t^{-\frac{1}{3}} \, dt = \left[ \frac{3}{2} t^{\frac{2}{3}} \right]_{-1}^{0} = \frac{3}{2} (0^{\frac{2}{3}} - (-1)^{\frac{2}{3}}) = \frac{3}{2} (0 - 1) = -\frac{3}{2} \]

Evaluating from \( 0 \) to \( 1 \):

\[ \int_{0}^{1} t^{-\frac{1}{3}} \, dt = \left[ \frac{3}{2} t^{\frac{2}{3}} \right]_{0}^{1} = \frac{3}{2} (1^{\frac{2}{3}} - 0^{\frac{2}{3}}) = \frac{3}{2} (1 - 0) = \frac{3}{2} \]

### Step 7: Combine Results
Now, combine both parts:

\[ I = -\frac{3}{2} + \frac{3}{2} = 0 \]

Thus, the value of the integral is:

\[ \boxed{0} \]