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

Options

A. `2sqrt(1+x)+log|(sqrt(1+x))/(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 \( \int \frac{\sqrt{1+x}}{x} \, dx \), we will use a substitution method. Here’s the step-by-step solution:

### Step 1: Substitution
Let \( t = \sqrt{1+x} \). Then, we have:
\[ 1 + x = t^2 \quad \Rightarrow \quad x = t^2 - 1 \]
Now, we differentiate both sides to find \( dx \):
\[ dx = 2t \, dt \]

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

### Step 3: Simplify the Integral
Now, we can simplify \( \frac{t^2}{t^2 - 1} \):
\[ \frac{t^2}{t^2 - 1} = 1 + \frac{1}{t^2 - 1} \]
Thus, the integral becomes:
\[ 2 \int \left( 1 + \frac{1}{t^2 - 1} \right) dt = 2 \int dt + 2 \int \frac{1}{t^2 - 1} \, dt \]

### Step 4: Integrate
Now we can integrate each term:
1. The first integral:
\[ 2 \int dt = 2t \]
2. The second integral:
\[ 2 \int \frac{1}{t^2 - 1} \, dt = 2 \cdot \frac{1}{2} \log \left| \frac{t-1}{t+1} \right| = \log \left| \frac{t-1}{t+1} \right| \]

Combining these results, we have:
\[ \int \frac{\sqrt{1+x}}{x} \, dx = 2t + \log \left| \frac{t-1}{t+1} \right| + C \]

### Step 5: Substitute Back
Now we substitute back \( t = \sqrt{1+x} \):
\[ = 2\sqrt{1+x} + \log \left| \frac{\sqrt{1+x}-1}{\sqrt{1+x}+1} \right| + C \]

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

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