Evaluate `int(log(sec^(-1)x)dx)/(xsqrt(x^(2)-1))dx`

Answer

To evaluate the integral

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

we will follow these steps:

### Step 1: Substitution
Let \( t = \sec^{-1} x \). Then, the derivative of \( t \) with respect to \( x \) is given by:

\[ \frac{dt}{dx} = \frac{1}{x \sqrt{x^2 - 1}} \implies dx = x \sqrt{x^2 - 1} \, dt. \]

### Step 2: Change of Variables
Substituting \( t = \sec^{-1} x \) into the integral, we have:

\[ I = \int \log(t) \, dt. \]

### Step 3: Integration by Parts
To integrate \( \log(t) \), we will use integration by parts. We set:

- \( u = \log(t) \) and \( dv = dt \)
- Then, \( du = \frac{1}{t} dt \) and \( v = t \)

Using the integration by parts formula:

\[ \int u \, dv = uv - \int v \, du, \]

we get:

\[ I = t \log(t) - \int t \cdot \frac{1}{t} \, dt = t \log(t) - \int dt. \]

### Step 4: Evaluate the Integral
The integral \( \int dt \) is simply \( t \). Therefore, we have:

\[ I = t \log(t) - t + C, \]

where \( C \) is the constant of integration.

### Step 5: Substitute Back
Now we substitute back \( t = \sec^{-1} x \):

\[ I = \sec^{-1} x \log(\sec^{-1} x) - \sec^{-1} x + C. \]

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

\[ I = \sec^{-1} x \log(\sec^{-1} x) - \sec^{-1} x + C. \]

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