`int(sqrt(1-x^(2))-x)/(sqrt(1-x^(2))(1+xsqrt(1-x^(2))))dx` is

Options

A. `2tan^(-1)(x+sqrt(1-x^(2)))+c`
B. `tan^(-1)(x+sqrt(1-x^(2)))+c`
C. `2tan^(-1)(x-sqrt(1-x^(2)))+c`
D. `2cot^(-1)(x+sqrt(1-x^(2)))+c`

Explanation

To solve the integral

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

we will use a trigonometric substitution. Let's proceed step by step.

### Step 1: Trigonometric Substitution
Let \( x = \sin \theta \). Then, we have:

\[ dx = \cos \theta \, d\theta. \]

Also, we can express \( \sqrt{1 - x^2} \) as:

\[ \sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta. \]

### Step 2: Substitute in the Integral
Substituting \( x \) and \( dx \) into the integral, we get:

\[ I = \int \frac{\cos \theta - \sin \theta}{\cos \theta(1 + \sin \theta \cos \theta)} \cos \theta \, d\theta. \]

This simplifies to:

\[ I = \int \frac{\cos^2 \theta - \sin \theta \cos \theta}{\cos \theta(1 + \sin \theta \cos \theta)} \, d\theta. \]

### Step 3: Simplify the Integral
Now, we can simplify the integral:

\[ I = \int \frac{\cos^2 \theta - \sin \theta \cos \theta}{1 + \sin \theta \cos \theta} \, d\theta. \]

### Step 4: Factor and Rewrite
We can factor out \( \cos \theta \):

\[ I = \int \frac{\cos \theta (\cos \theta - \sin \theta)}{1 + \sin \theta \cos \theta} \, d\theta. \]

### Step 5: Further Simplification
Now, let's multiply the numerator and denominator by 2:

\[ I = \int \frac{2(\cos \theta - \sin \theta)}{2 + 2\sin \theta \cos \theta} \, d\theta. \]

### Step 6: Recognize the Form
Notice that the denominator can be rewritten as:

\[ 2 + 2\sin \theta \cos \theta = 2(1 + \sin \theta \cos \theta). \]

Thus, we can simplify \( I \):

\[ I = \int \frac{\cos \theta - \sin \theta}{1 + \sin \theta \cos \theta} \, d\theta. \]

### Step 7: Substitution
Let \( t = \cos \theta + \sin \theta \). Then, we have:

\[ dt = (\cos \theta - \sin \theta) \, d\theta. \]

### Step 8: Change of Variables
Now, substituting \( t \) into the integral gives us:

\[ I = \int \frac{1}{1 + t^2} \, dt. \]

### Step 9: Solve the Integral
The integral of \( \frac{1}{1 + t^2} \) is:

\[ \int \frac{1}{1 + t^2} \, dt = \tan^{-1}(t) + C. \]

### Step 10: Back Substitute
Substituting back for \( t \):

\[ I = \tan^{-1}(\cos \theta + \sin \theta) + C. \]

### Step 11: Final Back Substitution
Using the original substitution \( x = \sin \theta \):

\[ \cos \theta = \sqrt{1 - x^2}, \quad \sin \theta = x. \]

Thus, we have:

\[ I = \tan^{-1}(\sqrt{1 - x^2} + x) + C. \]

### Final Answer:
\[ \int \frac{\sqrt{1 - x^2} - x}{\sqrt{1 - x^2}(1 + x\sqrt{1 - x^2})} \, dx = \tan^{-1}(\sqrt{1 - x^2} + x) + C. \]