Evaluate the following definite integrals:
`int_(0)^(1)x sqrt((1-x^(2))/(1+x^(2)))dx`

Answer

To evaluate the definite integral

\[ I = \int_{0}^{1} x \sqrt{\frac{1 - x^2}{1 + x^2}} \, dx, \]

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

### Step 1: Substitution
Let \( x^2 = t \). Then, we have:

\[ dx = \frac{1}{2\sqrt{t}} \, dt. \]

Now we need to change the limits of integration. When \( x = 0 \), \( t = 0 \) and when \( x = 1 \), \( t = 1 \). Thus, the integral becomes:

\[ I = \int_{0}^{1} \sqrt{t} \sqrt{\frac{1 - t}{1 + t}} \cdot \frac{1}{2\sqrt{t}} \, dt = \frac{1}{2} \int_{0}^{1} \sqrt{\frac{1 - t}{1 + t}} \, dt. \]

### Step 2: Simplifying the Integral
Now we can rewrite the integral:

\[ I = \frac{1}{2} \int_{0}^{1} \sqrt{\frac{1 - t}{1 + t}} \, dt. \]

### Step 3: Further Simplification
Next, we can multiply and divide the integrand by \(\sqrt{1 - t}\):

\[ I = \frac{1}{2} \int_{0}^{1} \frac{1 - t}{\sqrt{1 - t^2}} \, dt. \]

This can be split into two integrals:

\[ I = \frac{1}{2} \left( \int_{0}^{1} \frac{1}{\sqrt{1 - t^2}} \, dt - \int_{0}^{1} \frac{t}{\sqrt{1 - t^2}} \, dt \right). \]

### Step 4: Evaluating the Integrals
1. The first integral:

\[ \int_{0}^{1} \frac{1}{\sqrt{1 - t^2}} \, dt = \frac{\pi}{2}. \]

2. The second integral can be evaluated using the substitution \( t = \sin(\theta) \), \( dt = \cos(\theta) d\theta \):

\[ \int_{0}^{1} \frac{t}{\sqrt{1 - t^2}} \, dt = \int_{0}^{\frac{\pi}{2}} \sin(\theta) d\theta = -\cos(\theta) \bigg|_{0}^{\frac{\pi}{2}} = 1. \]

### Step 5: Combine Results
Putting it all together, we have:

\[ I = \frac{1}{2} \left( \frac{\pi}{2} - 1 \right). \]

Thus, the final result is:

\[ I = \frac{\pi}{4} - \frac{1}{2}. \]

### Final Answer
\[ \int_{0}^{1} x \sqrt{\frac{1 - x^2}{1 + x^2}} \, dx = \frac{\pi}{4} - \frac{1}{2}. \]