The value of `int (dx)/((1+sqrtx)(sqrt(x-x^2)))` is equal to

Options

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

Explanation

To solve the integral

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

we will use a substitution to simplify the expression.

### Step 1: Substitution
Let \( \sqrt{x} = \sin p \). Then, we have:

\[ x = \sin^2 p \quad \text{and} \quad dx = 2\sin p \cos p \, dp = \sin(2p) \, dp. \]

### Step 2: Rewrite the integral
Now, we need to rewrite the integral in terms of \( p \):

\[ \sqrt{x - x^2} = \sqrt{\sin^2 p - \sin^4 p} = \sqrt{\sin^2 p(1 - \sin^2 p)} = \sqrt{\sin^2 p \cos^2 p} = \sin p \cos p. \]

Substituting these into the integral, we get:

\[ I = \int \frac{\sin(2p) \, dp}{(1 + \sin p)(\sin p \cos p)}. \]

### Step 3: Simplifying the integral
This can be simplified to:

\[ I = \int \frac{2 \sin p \cos p \, dp}{(1 + \sin p)(\sin p \cos p)} = 2 \int \frac{dp}{1 + \sin p}. \]

### Step 4: Further simplification
To simplify \( \frac{1}{1 + \sin p} \), we can multiply the numerator and denominator by \( 1 - \sin p \):

\[ \frac{1 - \sin p}{(1 + \sin p)(1 - \sin p)} = \frac{1 - \sin p}{\cos^2 p}. \]

Thus, we have:

\[ I = 2 \int \frac{1 - \sin p}{\cos^2 p} \, dp = 2 \int \sec^2 p \, dp - 2 \int \frac{\sin p}{\cos^2 p} \, dp. \]

### Step 5: Integrate
The integrals can be computed as follows:

1. \( \int \sec^2 p \, dp = \tan p + C \).
2. \( \int \frac{\sin p}{\cos^2 p} \, dp = -\frac{1}{\cos p} + C \).

Thus, we have:

\[ I = 2 \left( \tan p + \frac{1}{\cos p} \right) + C. \]

### Step 6: Substitute back
Recall that \( \tan p = \frac{\sin p}{\cos p} \) and \( \sin p = \sqrt{x} \), \( \cos p = \sqrt{1 - x} \). Therefore:

\[ I = 2 \left( \frac{\sqrt{x}}{\sqrt{1 - x}} + \frac{1}{\sqrt{1 - x}} \right) + C = \frac{2\sqrt{x}}{\sqrt{1 - x}} - \frac{1}{\sqrt{1 - x}} + C. \]

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

\[ I = \frac{2\sqrt{x} - 1}{\sqrt{1 - x}} + C. \]