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

Options

A. `log(x+sqrt(x-1))+sin^(-1)sqrt((x-1)/(x))+C`
B. `log(x+sqrt(x-1))+C`
C. `log(x+sqrt(x-1))-(2)/(3)tan^(-1)((2sqrtx-1+1)/(sqrt(3)))+C`
D. none of these

Explanation

To solve the integral \( \int \frac{1}{x + \sqrt{x-1}} \, dx \), we will use a substitution method. Let's go through the steps:

### Step 1: Substitution
Let \( t^2 = x - 1 \). Then, we have:
\[ x = t^2 + 1 \]
Differentiating both sides gives:
\[ dx = 2t \, dt \]

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

### Step 3: Split the Integral
We can split the integral into two parts:
\[ \int \frac{2t + 1 - 1}{t^2 + t + 1} \, dt = \int \frac{2t + 1}{t^2 + t + 1} \, dt - \int \frac{1}{t^2 + t + 1} \, dt \]

### Step 4: First Integral
For the first integral \( \int \frac{2t + 1}{t^2 + t + 1} \, dt \), we can use the substitution \( u = t^2 + t + 1 \), then \( du = (2t + 1) \, dt \):
\[ \int \frac{du}{u} = \ln |u| + C_1 = \ln |t^2 + t + 1| + C_1 \]

### Step 5: Second Integral
For the second integral \( \int \frac{1}{t^2 + t + 1} \, dt \), we can complete the square:
\[ t^2 + t + 1 = \left(t + \frac{1}{2}\right)^2 + \frac{3}{4} \]
Thus, we have:
\[ \int \frac{1}{\left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, dt \]
This integral can be solved using the formula:
\[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \]
Applying this gives:
\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t + 1}{\sqrt{3}}\right) + C_2 \]

### Step 6: Combine Results
Combining the results from both integrals, we have:
\[ \int \frac{1}{x + \sqrt{x-1}} \, dx = \ln |t^2 + t + 1| - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t + 1}{\sqrt{3}}\right) + C \]

### Step 7: Back Substitute
Now, substitute back \( t = \sqrt{x - 1} \):
\[ \int \frac{1}{x + \sqrt{x-1}} \, dx = \ln |(\sqrt{x - 1})^2 + \sqrt{x - 1} + 1| - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2\sqrt{x - 1} + 1}{\sqrt{3}}\right) + C \]
This simplifies to:
\[ \ln |x + \sqrt{x - 1}| - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2\sqrt{x - 1} + 1}{\sqrt{3}}\right) + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{1}{x + \sqrt{x-1}} \, dx = \ln |x + \sqrt{x - 1}| - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2\sqrt{x - 1} + 1}{\sqrt{3}}\right) + C \]