Evaluate `int x^(-2//3)(1+x^(1//3))^(1//2)dx`.

Answer

To evaluate the integral \( \int x^{-\frac{2}{3}} (1 + x^{\frac{1}{3}})^{\frac{1}{2}} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t^2 = 1 + x^{\frac{1}{3}} \).

### Step 2: Differentiate both sides
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(t^2) = 2t \frac{dt}{dx} \quad \text{and} \quad \frac{d}{dx}(1 + x^{\frac{1}{3}}) = \frac{1}{3} x^{-\frac{2}{3}} \]
Thus, we have:
\[ 2t \frac{dt}{dx} = \frac{1}{3} x^{-\frac{2}{3}} \]

### Step 3: Solve for \( dx \)
Rearranging gives:
\[ dx = \frac{6t}{x^{-\frac{2}{3}}} dt \]

### Step 4: Express \( x^{-\frac{2}{3}} \) in terms of \( t \)
From our substitution \( t^2 = 1 + x^{\frac{1}{3}} \), we can express \( x^{\frac{1}{3}} \) as:
\[ x^{\frac{1}{3}} = t^2 - 1 \quad \Rightarrow \quad x = (t^2 - 1)^3 \]
Thus,
\[ x^{-\frac{2}{3}} = \frac{1}{(t^2 - 1)^2} \]

### Step 5: Substitute back into the integral
Now substituting back into the integral:
\[ \int x^{-\frac{2}{3}} (1 + x^{\frac{1}{3}})^{\frac{1}{2}} \, dx = \int \frac{1}{(t^2 - 1)^2} t \cdot \frac{6t}{x^{-\frac{2}{3}}} dt \]
This simplifies to:
\[ 6 \int t^2 dt \]

### Step 6: Integrate
Now we integrate:
\[ 6 \int t^2 dt = 6 \cdot \frac{t^3}{3} + C = 2t^3 + C \]

### Step 7: Substitute back for \( t \)
Finally, substituting back for \( t \):
\[ t = \sqrt{1 + x^{\frac{1}{3}}} \]
Thus, we have:
\[ 2 \left( \sqrt{1 + x^{\frac{1}{3}}} \right)^3 + C = 2(1 + x^{\frac{1}{3}})^{\frac{3}{2}} + C \]

### Final Answer
The evaluated integral is:
\[ \int x^{-\frac{2}{3}} (1 + x^{\frac{1}{3}})^{\frac{1}{2}} \, dx = 2(1 + x^{\frac{1}{3}})^{\frac{3}{2}} + C \]