The value of `int(sqrt(1+x))/(x)dx` , is

Options

A. `2sqrt(1+x)+log|(sqrt(1+x)-1)/(sqrt(1+x)+1)|+C`
B. `2sqrt(1+x)+C`
C. `log_(e)|(sqrt(1+)x-1)/(sqrt(1+x+1))|+C`
D. `(sqrt(1+x)-1)/(sqrt(1+x)+1)+C`

Explanation

To solve the integral \( I = \int \frac{\sqrt{1+x}}{x} \, dx \), we will use a substitution method.

### Step-by-Step Solution:

1. **Substitution:**
Let \( t^2 = 1 + x \). Then, we can express \( x \) in terms of \( t \):
\[ x = t^2 - 1 \]
Now, differentiate both sides to find \( dx \):
\[ dx = 2t \, dt \]

2. **Rewrite the Integral:**
Substitute \( x \) and \( dx \) into the integral:
\[ I = \int \frac{\sqrt{t^2}}{t^2 - 1} (2t \, dt) \]
Since \( \sqrt{t^2} = t \) (assuming \( t \geq 0 \)), we have:
\[ I = \int \frac{2t^2}{t^2 - 1} \, dt \]

3. **Separate the Integral:**
We can separate the integral into two parts:
\[ I = 2 \int \left( 1 + \frac{1}{t^2 - 1} \right) dt \]
This simplifies to:
\[ I = 2 \int dt + 2 \int \frac{1}{t^2 - 1} \, dt \]

4. **Integrate Each Part:**
The first integral is straightforward:
\[ 2 \int dt = 2t \]
The second integral can be solved using partial fraction decomposition:
\[ \frac{1}{t^2 - 1} = \frac{1}{(t-1)(t+1)} = \frac{1/2}{t-1} - \frac{1/2}{t+1} \]
Thus,
\[ \int \frac{1}{t^2 - 1} \, dt = \frac{1}{2} \ln |t-1| - \frac{1}{2} \ln |t+1| + C = \frac{1}{2} \ln \left| \frac{t-1}{t+1} \right| + C \]

5. **Combine Results:**
Now, substituting back into the integral:
\[ I = 2t + 2 \left( \frac{1}{2} \ln \left| \frac{t-1}{t+1} \right| \right) + C \]
Simplifying gives:
\[ I = 2t + \ln \left| \frac{t-1}{t+1} \right| + C \]

6. **Back Substitute \( t \):**
Recall that \( t = \sqrt{1+x} \):
\[ I = 2\sqrt{1+x} + \ln \left| \frac{\sqrt{1+x}-1}{\sqrt{1+x}+1} \right| + C \]

### Final Answer:
Thus, the value of the integral is:
\[ I = 2\sqrt{1+x} + \ln \left| \frac{\sqrt{1+x}-1}{\sqrt{1+x}+1} \right| + C \]