`int(x)/(sqrt(1+x^(2)))dx`

Answer

To solve the integral \(\int \frac{x}{\sqrt{1+x^2}} \, dx\), we can follow these steps:

### Step 1: Set up the integral
Let \(I = \int \frac{x}{\sqrt{1+x^2}} \, dx\).

### Step 2: Use substitution
We will use the substitution \(t = \sqrt{1+x^2}\). Then, squaring both sides gives us:
\[ t^2 = 1 + x^2 \implies x^2 = t^2 - 1 \]
Differentiating both sides with respect to \(x\):
\[ 2x \, dx = 2t \, dt \implies x \, dx = t \, dt \]

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

### Step 4: Integrate
The integral of \(dt\) is:
\[ I = t + C \]

### Step 5: Substitute back for \(t\)
Now, we substitute back \(t = \sqrt{1+x^2}\):
\[ I = \sqrt{1+x^2} + C \]

### Final Answer
Thus, the integral \(\int \frac{x}{\sqrt{1+x^2}} \, dx\) evaluates to:
\[ \int \frac{x}{\sqrt{1+x^2}} \, dx = \sqrt{1+x^2} + C \]

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