`int xsqrt(x^(4)+9) dx`

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int x \sqrt{x^4 + 9} \, dx \]
We can rewrite \( x^4 + 9 \) as \( (x^2)^2 + 3^2 \):
\[ I = \int x \sqrt{(x^2)^2 + 3^2} \, dx \]

### Step 2: Substitution
Let \( t = x^2 \). Then, differentiating both sides gives:
\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x} \]
Since \( x = \sqrt{t} \), we can substitute \( dx \) in terms of \( t \):
\[ dx = \frac{dt}{2\sqrt{t}} \]

### Step 3: Substitute in the Integral
Now, substituting \( x \) and \( dx \) into the integral, we have:
\[ I = \int \sqrt{t} \sqrt{t^2 + 9} \cdot \frac{dt}{2\sqrt{t}} = \frac{1}{2} \int \sqrt{t^2 + 9} \, dt \]

### Step 4: Use the Integral Formula
We can use the formula for the integral \( \int \sqrt{x^2 + a^2} \, dx \):
\[ \int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln |x + \sqrt{x^2 + a^2}| + C \]
In our case, \( a = 3 \) and we replace \( x \) with \( t \):
\[ \int \sqrt{t^2 + 9} \, dt = \frac{t}{2} \sqrt{t^2 + 9} + \frac{9}{2} \ln |t + \sqrt{t^2 + 9}| + C \]

### Step 5: Substitute Back
Now we substitute back \( t = x^2 \):
\[ I = \frac{1}{2} \left( \frac{x^2}{2} \sqrt{x^4 + 9} + \frac{9}{2} \ln |x^2 + \sqrt{x^4 + 9}| \right) + C \]
This simplifies to:
\[ I = \frac{x^2}{4} \sqrt{x^4 + 9} + \frac{9}{4} \ln |x^2 + \sqrt{x^4 + 9}| + C \]

### Final Answer
Thus, the final answer is:
\[ \int x \sqrt{x^4 + 9} \, dx = \frac{x^2}{4} \sqrt{x^4 + 9} + \frac{9}{4} \ln |x^2 + \sqrt{x^4 + 9}| + C \]