Evaluate : `int (dx)/( (1 - x^(2)) sqrt( 1 + x^(2)))`

Answer

To evaluate the integral

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

we will use a trigonometric substitution. Let's go through the steps:

### Step 1: Substitution
Let \( x = \sin \theta \). Then, we differentiate to find \( dx \):
\[ dx = \cos \theta \, d\theta. \]

### Step 2: Substitute into the Integral
Now, substituting \( x \) and \( dx \) into the integral, we have:
\[ \int \frac{\cos \theta \, d\theta}{(1 - \sin^2 \theta) \sqrt{1 + \sin^2 \theta}}. \]
Using the identity \( 1 - \sin^2 \theta = \cos^2 \theta \), the integral becomes:
\[ \int \frac{\cos \theta \, d\theta}{\cos^2 \theta \sqrt{1 + \sin^2 \theta}}. \]

### Step 3: Simplify the Integral
Now, simplify the expression:
\[ \int \frac{d\theta}{\cos \theta \sqrt{1 + \sin^2 \theta}}. \]
Next, we need to simplify \( \sqrt{1 + \sin^2 \theta} \). We know that:
\[ 1 + \sin^2 \theta = 1 + \frac{1 - \cos(2\theta)}{2} = \frac{3 + \cos(2\theta)}{2}. \]
However, for our purposes, we can leave it as \( \sqrt{1 + \sin^2 \theta} \).

### Step 4: Further Simplification
Now, we can express \( \sqrt{1 + \sin^2 \theta} \) in terms of \( \cos \theta \):
\[ \sqrt{1 + \sin^2 \theta} = \sqrt{1 + 1 - \cos^2 \theta} = \sqrt{2 - \cos^2 \theta}. \]
Thus, our integral is now:
\[ \int \frac{d\theta}{\cos \theta \sqrt{1 + \sin^2 \theta}}. \]

### Step 5: Integrate
This integral can be evaluated using the known integral:
\[ \int \sec \theta \, d\theta = \ln |\sec \theta + \tan \theta| + C. \]
Thus, we have:
\[ \int \frac{d\theta}{\cos \theta \sqrt{1 + \sin^2 \theta}} = \ln |\sec \theta + \tan \theta| + C. \]

### Step 6: Back Substitute
Now we need to substitute back for \( \theta \):
Since \( x = \sin \theta \), we have \( \theta = \sin^{-1}(x) \). Therefore:
\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\sqrt{1 - \sin^2 \theta}} = \frac{1}{\sqrt{1 - x^2}}, \]
and
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{x}{\sqrt{1 - x^2}}. \]
Thus, we can write:
\[ \sec \theta + \tan \theta = \frac{1}{\sqrt{1 - x^2}} + \frac{x}{\sqrt{1 - x^2}} = \frac{1 + x}{\sqrt{1 - x^2}}. \]

### Final Answer
Putting it all together, we have:
\[ \int \frac{dx}{(1 - x^2) \sqrt{1 + x^2}} = \ln \left| \frac{1 + x}{\sqrt{1 - x^2}} \right| + C. \]