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

Explanation

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

### Step 1: Simplify the integrand
We start with the integral:
\[ \int \frac{dx}{x + \sqrt{x}} \]
We can factor out \( \sqrt{x} \) from the denominator:
\[ = \int \frac{dx}{\sqrt{x}(\sqrt{x} + 1)} \]
This can be rewritten as:
\[ = \int \frac{1}{\sqrt{x}} \cdot \frac{1}{\sqrt{x} + 1} \, dx \]

### Step 2: Substitute \( t = \sqrt{x} + 1 \)
Let \( t = \sqrt{x} + 1 \). Then, we can express \( \sqrt{x} \) in terms of \( t \):
\[ \sqrt{x} = t - 1 \]
Now, squaring both sides gives:
\[ x = (t - 1)^2 \]
Next, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = \frac{1}{2\sqrt{x}} \implies dx = 2\sqrt{x} \, dt \]
Substituting \( \sqrt{x} = t - 1 \) into this gives:
\[ dx = 2(t - 1) \, dt \]

### Step 3: Substitute into the integral
Now we substitute \( dx \) and the expressions for \( \sqrt{x} \) and \( x \) into the integral:
\[ \int \frac{dx}{\sqrt{x}(\sqrt{x} + 1)} = \int \frac{2(t - 1) \, dt}{(t - 1)(t)} = \int \frac{2 \, dt}{t} \]

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

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

### Final Answer
Thus, the final answer is:
\[ \int \frac{dx}{x + \sqrt{x}} = 2 \ln |\sqrt{x} + 1| + C \]

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