`int(sqrt(x-1))/(x sqrt(x+1))dx " is equal to "`

Options

A. `"Ln "|x-sqrt(x^(2)-1)|-tan^(-1)x+c`
B. `"In "|x+sqrt(x^(2)-1)|-tan^(-1)x+c`
C. `"In "|x-sqrt(x^(2)-1)|-sec^(-1)x+c`
D. `"In "|x+sqrt(x^(2)-1)|-sec^(-1)x+c`

Explanation

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

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

### Step 2: Multiply and Simplify
To simplify the integral, we can multiply the numerator and denominator by \( \sqrt{x-1} \):
\[ I = \int \frac{(\sqrt{x-1})^2}{x \sqrt{x+1} \sqrt{x-1}} \, dx = \int \frac{x-1}{x \sqrt{x^2 - 1}} \, dx \]
Here, we used the identity \( a^2 - b^2 = (a-b)(a+b) \).

### Step 3: Split the Integral
Now, we can split the integral into two parts:
\[ I = \int \left( \frac{x}{x \sqrt{x^2 - 1}} - \frac{1}{x \sqrt{x^2 - 1}} \right) \, dx \]
This simplifies to:
\[ I = \int \frac{1}{\sqrt{x^2 - 1}} \, dx - \int \frac{1}{x \sqrt{x^2 - 1}} \, dx \]

### Step 4: Integrate Each Part
1. The first integral \( \int \frac{1}{\sqrt{x^2 - 1}} \, dx \) is known to be:
\[ \ln |x + \sqrt{x^2 - 1}| + C_1 \]
2. The second integral \( \int \frac{1}{x \sqrt{x^2 - 1}} \, dx \) is:
\[ \sec^{-1}(x) + C_2 \]

### Step 5: Combine the Results
Combining both results, we have:
\[ I = \ln |x + \sqrt{x^2 - 1}| - \sec^{-1}(x) + C \]
where \( C = C_1 - C_2 \) is the constant of integration.

### Final Result
Thus, the final answer is:
\[ \int \frac{\sqrt{x-1}}{x \sqrt{x+1}} \, dx = \ln |x + \sqrt{x^2 - 1}| - \sec^{-1}(x) + C \]