`int sqrt (1 + x ^(2)). x dx = (1)/(3) (1 + x ^(2)) ^((3)/(2)) + c`

Answer

To solve the integral \( \int \sqrt{1 + x^2} \cdot x \, dx \) and verify the given statement, we will follow these steps:

### Step 1: Set up the integral
Let
\[ I = \int \sqrt{1 + x^2} \cdot x \, dx \]

### Step 2: Use substitution
We will use the substitution \( t^2 = 1 + x^2 \). Therefore, we differentiate both sides:
\[ 2t \, dt = 2x \, dx \implies x \, dx = t \, dt \]

### Step 3: Express \( \sqrt{1 + x^2} \) in terms of \( t \)
From our substitution, we have:
\[ \sqrt{1 + x^2} = t \]

### Step 4: Substitute into the integral
Now we can rewrite the integral \( I \):
\[ I = \int t \cdot (t \, dt) = \int t^2 \, dt \]

### Step 5: Integrate
The integral of \( t^2 \) is:
\[ \int t^2 \, dt = \frac{t^3}{3} + C \]

### Step 6: Substitute back for \( t \)
Now we substitute back \( t = \sqrt{1 + x^2} \):
\[ I = \frac{(\sqrt{1 + x^2})^3}{3} + C = \frac{(1 + x^2)^{3/2}}{3} + C \]

### Conclusion
Thus, we have shown that:
\[ \int \sqrt{1 + x^2} \cdot x \, dx = \frac{1}{3} (1 + x^2)^{3/2} + C \]
This confirms that the given statement is true.

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