`int(xlnx)/(x^(2)-1)^(3//2)` dx equals

Options

A. arc `secx-(lnx)/sqrt(x^(2)-1)+C`
B. `sec^(-1)x+(lnx)/sqrt(x^(2)-1)+C`
C. `cos^(-1)x-(lnx)/sqrt(x^(2)-1)+C`
D. `secx-(lnx)/sqrt(x^(2)-1)+C`

Answer

To solve the integral

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

we will use integration by parts. Let's denote:

- \( u = \ln x \)
- \( dv = \frac{x}{(x^2 - 1)^{3/2}} \, dx \)

Next, we need to find \( du \) and \( v \):

1. **Finding \( du \)**:
\[ du = \frac{1}{x} \, dx \]

2. **Finding \( v \)**:
To find \( v \), we need to integrate \( dv \):
\[ v = \int \frac{x}{(x^2 - 1)^{3/2}} \, dx \]
We can use the substitution \( t = x^2 - 1 \), which gives:
\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x} \]
Thus, we have:
\[ v = \int \frac{x}{t^{3/2}} \cdot \frac{dt}{2x} = \frac{1}{2} \int t^{-3/2} \, dt \]
The integral of \( t^{-3/2} \) is:
\[ \int t^{-3/2} \, dt = -2 t^{-1/2} + C = -\frac{2}{\sqrt{t}} + C \]
Therefore:
\[ v = -\frac{1}{\sqrt{x^2 - 1}} \]

Now we can apply the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values:
\[ I = \ln x \left(-\frac{1}{\sqrt{x^2 - 1}}\right) - \int \left(-\frac{1}{\sqrt{x^2 - 1}}\right) \left(\frac{1}{x}\right) \, dx \]
This simplifies to:
\[ I = -\frac{\ln x}{\sqrt{x^2 - 1}} + \int \frac{1}{x \sqrt{x^2 - 1}} \, dx \]

Next, we need to solve the integral \( \int \frac{1}{x \sqrt{x^2 - 1}} \, dx \). We can use the substitution \( x = \sec \theta \), which gives:
\[ dx = \sec \theta \tan \theta \, d\theta \]
Thus:
\[ \int \frac{1}{\sec \theta \sqrt{\sec^2 \theta - 1}} \sec \theta \tan \theta \, d\theta = \int \frac{\tan \theta}{\tan \theta} \, d\theta = \int d\theta = \theta + C \]
Since \( \theta = \sec^{-1}(x) \), we have:
\[ \int \frac{1}{x \sqrt{x^2 - 1}} \, dx = \sec^{-1}(x) \]

Putting everything together:
\[ I = -\frac{\ln x}{\sqrt{x^2 - 1}} + \sec^{-1}(x) + C \]

Thus, the final answer is:
\[ \int \frac{x \ln x}{(x^2 - 1)^{3/2}} \, dx = -\frac{\ln x}{\sqrt{x^2 - 1}} + \sec^{-1}(x) + C \]

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