What is the value of `int(sqrt(x)+x)^(-1)` ?

Options

A. `ln(x+sqrt(x))+c`
B. `2ln(1+sqrt(x))+c`
C. `2ln(x+sqrt(x))+c`
D. `2ln(x-sqrt(x))+c`

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{1}{\sqrt{x} + x} \, dx \]

### Step 2: Factor out \( \sqrt{x} \)
We can factor \( \sqrt{x} \) out of the denominator:
\[ \int \frac{1}{\sqrt{x}(1 + \sqrt{x})} \, dx \]

### Step 3: Substitution
Let \( t = 1 + \sqrt{x} \). Then, we differentiate \( t \) with respect to \( x \):
\[ \sqrt{x} = t - 1 \quad \text{and} \quad x = (t - 1)^2 \]
Now, we need to find \( dx \):
\[ dx = 2(t - 1) \, dt \]

### Step 4: Substitute in the Integral
Substituting \( \sqrt{x} \) and \( dx \) into the integral gives:
\[ \int \frac{2(t - 1)}{(t - 1)(1 + \sqrt{x})} \, dt = \int \frac{2 \, dt}{t} \]

### Step 5: Integrate
Now we can integrate:
\[ \int \frac{2}{t} \, dt = 2 \ln |t| + C \]

### Step 6: Back Substitute
Now we substitute back \( t = 1 + \sqrt{x} \):
\[ 2 \ln |1 + \sqrt{x}| + C \]

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

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