To evaluate the integral
\[
I = \int \frac{x \log x}{(x^2 - 1)^{3/2}} \, dx,
\]
we will use integration by parts. Let's denote:
- \( u = \log x \)
- \( dv = \frac{x}{(x^2 - 1)^{3/2}} \, dx \)
### Step 1: Differentiate \( u \) and integrate \( dv \)
First, we differentiate \( u \):
\[
du = \frac{1}{x} \, dx.
\]
Next, we need to integrate \( dv \):
To integrate \( dv = \frac{x}{(x^2 - 1)^{3/2}} \, dx \), we can use a substitution. Let:
\[
t = x^2 - 1 \implies dt = 2x \, dx \implies x \, dx = \frac{dt}{2}.
\]
Then, we rewrite the integral:
\[
\int \frac{x}{(x^2 - 1)^{3/2}} \, dx = \int \frac{1}{t^{3/2}} \cdot \frac{dt}{2} = \frac{1}{2} \int t^{-3/2} \, dt.
\]
### Step 2: Integrate \( t^{-3/2} \)
The integral of \( t^{-3/2} \) is:
\[
\int t^{-3/2} \, dt = -2 t^{-1/2} + C = -\frac{2}{\sqrt{t}} + C.
\]
Substituting back \( t = x^2 - 1 \):
\[
\int \frac{x}{(x^2 - 1)^{3/2}} \, dx = -\frac{1}{\sqrt{x^2 - 1}} + C.
\]
### Step 3: Apply integration by parts
Now we can apply integration by parts:
\[
I = u \cdot v - \int v \, du.
\]
Substituting \( u \) and \( v \):
\[
I = \log x \left(-\frac{1}{\sqrt{x^2 - 1}}\right) - \int \left(-\frac{1}{\sqrt{x^2 - 1}}\right) \left(\frac{1}{x}\right) \, dx.
\]
This simplifies to:
\[
I = -\frac{\log x}{\sqrt{x^2 - 1}} + \int \frac{1}{x \sqrt{x^2 - 1}} \, dx.
\]
### Step 4: Evaluate the remaining integral
The integral
\[
\int \frac{1}{x \sqrt{x^2 - 1}} \, dx
\]
can be solved by the substitution \( x = \sec \theta \), where \( dx = \sec \theta \tan \theta \, d\theta \). Thus, we have:
\[
\sqrt{x^2 - 1} = \sqrt{\sec^2 \theta - 1} = \tan \theta.
\]
So the integral becomes:
\[
\int \frac{\sec \theta \tan \theta}{\sec \theta \tan \theta} \, d\theta = \int d\theta = \theta + C = \sec^{-1}(x) + C.
\]
### Step 5: Combine results
Putting everything together, we have:
\[
I = -\frac{\log x}{\sqrt{x^2 - 1}} + \sec^{-1}(x) + C.
\]
Thus, the final answer is:
\[
\int \frac{x \log x}{(x^2 - 1)^{3/2}} \, dx = -\frac{\log x}{\sqrt{x^2 - 1}} + \sec^{-1}(x) + C.
\]
