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

Explanation

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

### Step 1: Substitution
Let \(x = t^2\). Then, \(dx = 2t \, dt\) and \(\sqrt{x} = t\).

### Step 2: Rewrite the Integral
Substituting \(x\) and \(dx\) into the integral, we get:
\[ \int \frac{t^2}{t + 1} \cdot 2t \, dt = 2 \int \frac{t^3}{t + 1} \, dt \]

### Step 3: Simplify the Integrand
Now, we can simplify \(\frac{t^3}{t + 1}\) using polynomial long division:
\[ \frac{t^3}{t + 1} = t^2 - t + 1 - \frac{1}{t + 1} \]
Thus, we can rewrite the integral as:
\[ 2 \int \left(t^2 - t + 1 - \frac{1}{t + 1}\right) \, dt \]

### Step 4: Integrate Each Term
Now we can integrate each term separately:
\[ 2 \left( \int t^2 \, dt - \int t \, dt + \int 1 \, dt - \int \frac{1}{t + 1} \, dt \right) \]
Calculating each integral:
- \(\int t^2 \, dt = \frac{t^3}{3}\)
- \(\int t \, dt = \frac{t^2}{2}\)
- \(\int 1 \, dt = t\)
- \(\int \frac{1}{t + 1} \, dt = \ln |t + 1|\)

Putting it all together:
\[ 2 \left( \frac{t^3}{3} - \frac{t^2}{2} + t - \ln |t + 1| \right) + C \]

### Step 5: Substitute Back
Now, substitute back \(t = \sqrt{x}\):
\[ = \frac{2(\sqrt{x})^3}{3} - 2 \cdot \frac{(\sqrt{x})^2}{2} + 2\sqrt{x} - 2\ln |\sqrt{x} + 1| + C \]
This simplifies to:
\[ = \frac{2x^{3/2}}{3} - x + 2\sqrt{x} - 2\ln (\sqrt{x} + 1) + C \]

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

---