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

Options

A. `- sqrt2 - 1`
B. `sqrt2 + 1`
C. `- sqrt2 + 1`
D. `sqrt2 - 1`

Explanation

To solve the integral \( I = \int_{0}^{1} \frac{x}{\sqrt{1 + x^2}} \, dx \), we will use substitution. Here are the steps:

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

### Step 2: Change the limits of integration
When \( x = 0 \):
\[ t = 1 + 0^2 = 1 \]
When \( x = 1 \):
\[ t = 1 + 1^2 = 2 \]

### Step 3: Rewrite the integral
Now we can rewrite the integral in terms of \( t \):
\[ I = \int_{1}^{2} \frac{1}{\sqrt{t}} \cdot \frac{dt}{2} \]
This simplifies to:
\[ I = \frac{1}{2} \int_{1}^{2} t^{-1/2} \, dt \]

### Step 4: Integrate
Now, we integrate \( t^{-1/2} \):
\[ \int t^{-1/2} \, dt = 2t^{1/2} + C \]
Thus,
\[ I = \frac{1}{2} \left[ 2t^{1/2} \right]_{1}^{2} \]
This simplifies to:
\[ I = \left[ t^{1/2} \right]_{1}^{2} \]

### Step 5: Evaluate the definite integral
Now, evaluate the limits:
\[ I = \left[ \sqrt{2} - \sqrt{1} \right] = \sqrt{2} - 1 \]

### Final Answer
Thus, the value of the integral is:
\[ I = \sqrt{2} - 1 \]

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