Evaluate `int(dx)/((1-x^(3))^(1//3))dx`.

Answer

To evaluate the integral

\[ I = \int \frac{dx}{(1 - x^3)^{1/3}}, \]

we will use substitution and integration techniques.

### Step 1: Substitution
Let us substitute \( x = \frac{1}{t} \). Then, we have:

\[ dx = -\frac{1}{t^2} dt. \]

### Step 2: Change of Variables
Substituting \( x = \frac{1}{t} \) into the integral, we get:

\[ I = \int \frac{-\frac{1}{t^2} dt}{\left(1 - \left(\frac{1}{t}\right)^3\right)^{1/3}} = -\int \frac{dt}{t^2 \left(1 - \frac{1}{t^3}\right)^{1/3}}. \]

### Step 3: Simplifying the Denominator
The denominator simplifies as follows:

\[ 1 - \frac{1}{t^3} = \frac{t^3 - 1}{t^3}. \]

Thus, we have:

\[ \left(1 - \frac{1}{t^3}\right)^{1/3} = \left(\frac{t^3 - 1}{t^3}\right)^{1/3} = \frac{(t^3 - 1)^{1/3}}{t}. \]

### Step 4: Substitute Back into the Integral
Now substituting this back into the integral, we get:

\[ I = -\int \frac{t \, dt}{t^2 (t^3 - 1)^{1/3}} = -\int \frac{dt}{t (t^3 - 1)^{1/3}}. \]

### Step 5: Further Substitution
Let \( u = t^3 - 1 \). Then, \( du = 3t^2 dt \) or \( dt = \frac{du}{3t^2} \). Since \( t^2 = (u + 1)^{2/3} \), we can write:

\[ dt = \frac{du}{3(u + 1)^{2/3}}. \]

### Step 6: Substitute \( t \) in Terms of \( u \)
The integral becomes:

\[ I = -\int \frac{1}{(u + 1)^{1/3}} \cdot \frac{du}{3(u + 1)^{2/3}} = -\frac{1}{3} \int \frac{du}{(u + 1)}. \]

### Step 7: Integrate
The integral simplifies to:

\[ I = -\frac{1}{3} \ln |u + 1| + C. \]

### Step 8: Substitute Back for \( u \)
Substituting back for \( u \):

\[ I = -\frac{1}{3} \ln |t^3| + C = -\frac{1}{3} \ln |(1 - x^3)^{1/3}| + C. \]

### Final Result
Thus, the final result is:

\[ I = -\frac{1}{3} \ln |1 - x^3| + C. \]