Evaluate `intsqrt((x)/(1-x^3))dx`

Answer

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

### Step 1: Rewrite the Integral
We can rewrite the integral as:
\[ \int \frac{\sqrt{x}}{\sqrt{1-x^3}} \, dx \]

### Step 2: Substitution
Let us use the substitution \(t = x^{3/2}\). Then, we have:
\[ x = t^{2/3} \]
Differentiating both sides gives:
\[ dx = \frac{2}{3} t^{-1/3} \, dt \]

### Step 3: Substitute in the Integral
Now, we substitute \(x\) and \(dx\) in the integral:
\[ \sqrt{x} = \sqrt{t^{2/3}} = t^{1/3} \]
\[ 1 - x^3 = 1 - (t^{2/3})^3 = 1 - t^2 \]
Thus, the integral becomes:
\[ \int \frac{t^{1/3}}{\sqrt{1-t^2}} \cdot \frac{2}{3} t^{-1/3} \, dt \]
This simplifies to:
\[ \frac{2}{3} \int \frac{1}{\sqrt{1-t^2}} \, dt \]

### Step 4: Evaluate the Integral
The integral \(\int \frac{1}{\sqrt{1-t^2}} \, dt\) is a standard integral that evaluates to \(\sin^{-1}(t)\). Therefore, we have:
\[ \frac{2}{3} \sin^{-1}(t) + C \]

### Step 5: Substitute Back
Now, we substitute back \(t = x^{3/2}\):
\[ \frac{2}{3} \sin^{-1}(x^{3/2}) + C \]

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