`int(1)/((x-1)sqrt(x^(2)-1))dx` equals

Options

A. `-sqrt((x-1)/(x+1))+C`
B. `sqrt((x-1)/(x+1))+C`
C. `sqrt((x+1)/(x-1))+C`
D. `-sqrt((x+1)/(x-1))+C`

Explanation

To solve the integral

\[ I = \int \frac{1}{(x-1) \sqrt{x^2 - 1}} \, dx, \]

we will follow a series of steps:

### Step 1: Rewrite the integral
We start by rewriting the integral in a more manageable form. The expression \(\sqrt{x^2 - 1}\) can be factored as:

\[ \sqrt{x^2 - 1} = \sqrt{(x-1)(x+1)}. \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{1}{(x-1) \sqrt{(x-1)(x+1)}} \, dx = \int \frac{1}{(x-1)^{3/2} \sqrt{x+1}} \, dx. \]

### Step 2: Multiply and divide by \((x+1)^{3/2}\)
To simplify the integral further, we multiply and divide by \((x+1)^{3/2}\):

\[ I = \int \frac{(x+1)^{3/2}}{(x-1)^{3/2} (x+1)} \, dx = \int \frac{(x+1)^{1/2}}{(x-1)^{3/2}} \, dx. \]

### Step 3: Substitution
Now, we will use the substitution:

\[ t = \frac{x-1}{x+1}. \]

This implies:

\[ x = \frac{1+t}{1-t}. \]

Differentiating \(x\) with respect to \(t\):

\[ dx = \frac{2}{(1-t)^2} \, dt. \]

### Step 4: Change of variables in the integral
We need to express \((x+1)\) and \((x-1)\) in terms of \(t\):

\[ x-1 = \frac{2t}{1-t}, \quad x+1 = \frac{2}{1-t}. \]

Now substituting these into the integral gives:

\[ I = \int \frac{\left(\frac{2}{1-t}\right)^{1/2}}{\left(\frac{2t}{1-t}\right)^{3/2}} \cdot \frac{2}{(1-t)^2} \, dt. \]

### Step 5: Simplifying the integral
After simplification, we have:

\[ I = \int \frac{2^{1/2}}{(2t)^{3/2}} \cdot \frac{2}{(1-t)^2} \, dt = \int \frac{1}{t^{3/2}(1-t)^2} \, dt. \]

### Step 6: Integrating
Now we can integrate:

\[ I = -2t^{-1/2} + C. \]

### Step 7: Back substitution
Substituting back for \(t\):

\[ t = \frac{x-1}{x+1} \implies I = -2 \left(\frac{x+1}{x-1}\right)^{1/2} + C. \]

### Final Result
Thus, the final result for the integral is:

\[ I = -\sqrt{\frac{x+1}{x-1}} + C. \]