`int_(0)^(1) (x)/(sqrt(1+x^(2)))dx=?`

Answer

To solve the integral \( \int_{0}^{1} \frac{x}{\sqrt{1+x^2}} \, dx \), we will use a substitution method.

### Step-by-Step Solution:

1. **Substitution**: Let \( t = 1 + x^2 \). Then, we differentiate both sides with respect to \( x \):
\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x} \]

2. **Change of Variables**: We also need to express \( x \) in terms of \( t \):
\[ x^2 = t - 1 \quad \Rightarrow \quad x = \sqrt{t - 1} \]
Now, we need to change the limits of integration. When \( x = 0 \):
\[ t = 1 + 0^2 = 1 \]
When \( x = 1 \):
\[ t = 1 + 1^2 = 2 \]
Therefore, the limits change from \( x: 0 \to 1 \) to \( t: 1 \to 2 \).

3. **Substituting in the Integral**: Substitute \( x \) and \( dx \) into the integral:
\[ \int_{0}^{1} \frac{x}{\sqrt{1+x^2}} \, dx = \int_{1}^{2} \frac{\sqrt{t-1}}{\sqrt{t}} \cdot \frac{dt}{2\sqrt{t-1}} = \int_{1}^{2} \frac{1}{2\sqrt{t}} \, dt \]

4. **Simplifying the Integral**: The integral simplifies to:
\[ \frac{1}{2} \int_{1}^{2} t^{-1/2} \, dt \]

5. **Integrating**: The integral of \( t^{-1/2} \) is:
\[ \int t^{-1/2} \, dt = 2t^{1/2} + C \]
Therefore,
\[ \frac{1}{2} \left[ 2t^{1/2} \right]_{1}^{2} = \left[ t^{1/2} \right]_{1}^{2} \]

6. **Evaluating the Limits**: Now we evaluate the limits:
\[ = \sqrt{2} - \sqrt{1} = \sqrt{2} - 1 \]

### Final Answer:
Thus, the value of the integral is:
\[ \int_{0}^{1} \frac{x}{\sqrt{1+x^2}} \, dx = \sqrt{2} - 1 \]