`int(x)/(sqrt(9-x^(4)))dx`

Explanation

To solve the integral \( \int \frac{x}{\sqrt{9 - x^4}} \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = x^2 \). Then, the differential \( dt = 2x \, dx \) or \( dx = \frac{dt}{2x} \).

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we have:
\[ x = \sqrt{t} \quad \text{and} \quad x^4 = t^2 \]
Thus, the integral becomes:
\[ \int \frac{\sqrt{t}}{\sqrt{9 - t^2}} \cdot \frac{dt}{2\sqrt{t}} = \int \frac{1}{2\sqrt{9 - t^2}} \, dt \]

### Step 3: Simplify the Integral
Now, we can simplify the integral:
\[ \int \frac{1}{2\sqrt{9 - t^2}} \, dt \]
This can be factored out as:
\[ \frac{1}{2} \int \frac{1}{\sqrt{9 - t^2}} \, dt \]

### Step 4: Use the Standard Integral Formula
The integral \( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1} \left( \frac{x}{a} \right) + C \) applies here, where \( a = 3 \):
\[ \frac{1}{2} \cdot \sin^{-1} \left( \frac{t}{3} \right) + C \]

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

### Final Answer
Thus, the final answer is:
\[ \int \frac{x}{\sqrt{9 - x^4}} \, dx = \frac{1}{2} \sin^{-1} \left( \frac{x^2}{3} \right) + C \]

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