`int sqrt(x^2+1)/x dx`

Answer

To solve the integral \( \int \frac{\sqrt{x^2 + 1}}{x} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = \sqrt{x^2 + 1} \). Then, squaring both sides gives us:
\[ t^2 = x^2 + 1 \implies x^2 = t^2 - 1 \implies x = \sqrt{t^2 - 1} \]

### Step 2: Differentiate to find \( dx \)
Differentiating both sides with respect to \( t \):
\[ dx = \frac{d}{dt}(\sqrt{t^2 - 1}) = \frac{1}{2\sqrt{t^2 - 1}} \cdot 2t \, dt = \frac{t}{\sqrt{t^2 - 1}} \, dt \]

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

### Step 4: Simplify the integrand
We can rewrite the integrand:
\[ \frac{t^2}{t^2 - 1} = 1 + \frac{1}{t^2 - 1} \]
Thus, the integral becomes:
\[ \int \left( 1 + \frac{1}{t^2 - 1} \right) dt = \int 1 \, dt + \int \frac{1}{t^2 - 1} \, dt \]

### Step 5: Integrate each part
1. The integral of \( 1 \) is:
\[ \int 1 \, dt = t \]
2. For the integral \( \int \frac{1}{t^2 - 1} \, dt \), we can use the formula:
\[ \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C \]
Here, \( a = 1 \):
\[ \int \frac{1}{t^2 - 1} \, dt = \frac{1}{2} \log \left| \frac{t - 1}{t + 1} \right| \]

### Step 6: Combine results
Combining both parts, we have:
\[ \int \frac{\sqrt{x^2 + 1}}{x} \, dx = t + \frac{1}{2} \log \left| \frac{t - 1}{t + 1} \right| + C \]

### Step 7: Substitute back for \( t \)
Recall that \( t = \sqrt{x^2 + 1} \):
\[ \int \frac{\sqrt{x^2 + 1}}{x} \, dx = \sqrt{x^2 + 1} + \frac{1}{2} \log \left| \frac{\sqrt{x^2 + 1} - 1}{\sqrt{x^2 + 1} + 1} \right| + C \]

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