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

Options

A. `sqrt(x)-log|1+sqrt(x)|+C`
B. `sqrt(x)+log|1+sqrt(x)|+C`
C. `2sqrt(x)-2log|1+sqrt(x)|+C`
D. none of these

Explanation

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

### Step 1: Substitution
Let \( x = t^2 \). Then, we differentiate to find \( dx \):
\[ dx = 2t \, dt \]

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

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

### Step 4: Separate the Integral
We can separate the fraction:
\[ \frac{t}{1 + t} = 1 - \frac{1}{1 + t} \]
Thus, we rewrite the integral:
\[ 2 \int \left(1 - \frac{1}{1 + t}\right) dt = 2 \left( \int dt - \int \frac{1}{1 + t} \, dt \right) \]

### Step 5: Integrate
Now we can integrate each term:
\[ 2 \left( t - \ln |1 + t| \right) + C \]

### Step 6: Substitute Back
Recall that we made the substitution \( t = \sqrt{x} \). Thus, substituting back gives:
\[ 2 \left( \sqrt{x} - \ln |1 + \sqrt{x}| \right) + C \]

### Final Answer
The final result of the integral is:
\[ \int \frac{dx}{1 + \sqrt{x}} = 2\sqrt{x} - 2\ln |1 + \sqrt{x}| + C \]

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