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

Answer

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

### Step 1: Substitution
Let \( t = x^{\frac{1}{3}} \). Then, we differentiate both sides to find \( dx \):
\[ \frac{dt}{dx} = \frac{1}{3} x^{-\frac{2}{3}} \implies dx = 3t^2 \, dt \]

### Step 2: Express \( x \) in terms of \( t \)
From our substitution, we have:
\[ x = t^3 \quad \text{and thus} \quad x^{\frac{2}{3}} = t^2 \]

### Step 3: Substitute in the integral
Now we can substitute \( x \) and \( dx \) in the integral:
\[ I = \int (t^3)^{-\frac{2}{3}} (1 + (t^3)^{\frac{2}{3}})^{-1} (3t^2 \, dt) \]
This simplifies to:
\[ I = \int t^{-2} (1 + t^2)^{-1} (3t^2 \, dt) \]

### Step 4: Simplify the integral
The expression simplifies further:
\[ I = 3 \int \frac{t^2}{t^2(1 + t^2)} \, dt = 3 \int \frac{1}{1 + t^2} \, dt \]

### Step 5: Integrate
We know that:
\[ \int \frac{1}{1 + t^2} \, dt = \tan^{-1}(t) + C \]
Thus, we have:
\[ I = 3 \tan^{-1}(t) + C \]

### Step 6: Back substitute for \( t \)
Now substituting back \( t = x^{\frac{1}{3}} \):
\[ I = 3 \tan^{-1}(x^{\frac{1}{3}}) + C \]

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