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

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{1}{x + \sqrt{x}} \, dx \]
We can factor out \( \sqrt{x} \) from the denominator:
\[ = \int \frac{1}{\sqrt{x}(\sqrt{x} + 1)} \, dx \]

### Step 2: Substitute \( \sqrt{x} \)
Let \( t = \sqrt{x} \). Then, \( x = t^2 \) and \( dx = 2t \, dt \). Substituting these into the integral gives:
\[ = \int \frac{2t}{t^2 + t} \, dt \]

### Step 3: Simplify the Integral
Now, we can simplify the expression:
\[ = \int \frac{2t}{t(t + 1)} \, dt = \int \frac{2}{t + 1} \, dt \]

### Step 4: Integrate
Now we can integrate:
\[ = 2 \int \frac{1}{t + 1} \, dt = 2 \ln |t + 1| + C \]

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

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