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

Options

A. `"In "|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: Rationalization
We start by rationalizing the integrand. We multiply the numerator and the denominator by \( \sqrt{x-1} \):

\[ \int \frac{\sqrt{x-1}}{x \sqrt{x+1}} \, dx = \int \frac{\sqrt{x-1} \cdot \sqrt{x-1}}{x \sqrt{x+1} \cdot \sqrt{x-1}} \, dx = \int \frac{x-1}{x \sqrt{x^2 - 1}} \, dx \]

### Step 2: Simplifying the Integral
This can be rewritten as:

\[ \int \frac{1}{\sqrt{x^2 - 1}} \, dx - \int \frac{1}{x \sqrt{x^2 - 1}} \, dx \]

### Step 3: Using Integration Formulas
Now we can apply the known integration formulas. The integral \( \int \frac{1}{\sqrt{x^2 - 1}} \, dx \) is:

\[ \ln |x + \sqrt{x^2 - 1}| + C \]

And the integral \( \int \frac{1}{x \sqrt{x^2 - 1}} \, dx \) is:

\[ \sec^{-1}(x) + C \]

### Step 4: Combining the Results
Putting it all together, we have:

\[ \int \frac{\sqrt{x-1}}{x \sqrt{x+1}} \, dx = \ln |x + \sqrt{x^2 - 1}| - \sec^{-1}(x) + C \]

### Final Answer
Thus, the final result of the integral is:

\[ \int \frac{\sqrt{x-1}}{x \sqrt{x+1}} \, dx = \ln |x + \sqrt{x^2 - 1}| - \sec^{-1}(x) + C \]

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