Evaluate the following integrals
`int sqrt((1-x)/(1+x))dx`

Answer

To evaluate the integral \( \int \sqrt{\frac{1-x}{1+x}} \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rationalize the integrand**:
We can rewrite the integral as:
\[ \int \sqrt{\frac{1-x}{1+x}} \, dx = \int \frac{\sqrt{1-x}}{\sqrt{1+x}} \, dx \]
To simplify this, we can multiply the numerator and denominator by \( \sqrt{1-x} \):
\[ = \int \frac{(1-x)}{\sqrt{(1-x)(1+x)}} \, dx \]

2. **Simplify the expression**:
The denominator can be simplified using the identity \( \sqrt{(1-x)(1+x)} = \sqrt{1-x^2} \):
\[ = \int \frac{1-x}{\sqrt{1-x^2}} \, dx \]

3. **Split the integral**:
We can split the integral into two parts:
\[ = \int \frac{1}{\sqrt{1-x^2}} \, dx - \int \frac{x}{\sqrt{1-x^2}} \, dx \]

4. **Evaluate the first integral**:
The first integral \( \int \frac{1}{\sqrt{1-x^2}} \, dx \) is a standard integral that evaluates to:
\[ \sin^{-1}(x) \]

5. **Evaluate the second integral**:
For the second integral, we can use the substitution \( u = 1-x^2 \), which gives \( du = -2x \, dx \) or \( dx = -\frac{du}{2x} \). However, a simpler approach is to recognize that:
\[ \int \frac{x}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2} \]
Therefore, we have:
\[ -\int \frac{x}{\sqrt{1-x^2}} \, dx = \sqrt{1-x^2} \]

6. **Combine the results**:
Combining both results, we get:
\[ \int \sqrt{\frac{1-x}{1+x}} \, dx = \sin^{-1}(x) + \sqrt{1-x^2} + C \]

### Final Answer:
Thus, the evaluated integral is:
\[ \int \sqrt{\frac{1-x}{1+x}} \, dx = \sin^{-1}(x) + \sqrt{1-x^2} + C \]