If `f'(x)=(1)/((1+x^(2))^(3//2))` and `f(0)=0,` then `f(1)` is equal to :

Options

A. `sqrt2`
B. `-1/(sqrt(2))`
C. `1/(sqrt(2))`
D. None of these

Explanation

To solve the problem, we need to find the function \( f(x) \) given its derivative \( f'(x) = \frac{1}{(1+x^2)^{3/2}} \) and the condition \( f(0) = 0 \). We will then evaluate \( f(1) \).

### Step-by-Step Solution:

1. **Integrate \( f'(x) \)**:
\[ f(x) = \int f'(x) \, dx = \int \frac{1}{(1+x^2)^{3/2}} \, dx \]

2. **Use a trigonometric substitution**:
Let \( x = \tan(\theta) \). Then, \( dx = \sec^2(\theta) \, d\theta \) and \( 1 + x^2 = \sec^2(\theta) \).
Substituting these into the integral gives:
\[ f(x) = \int \frac{\sec^2(\theta)}{(\sec^2(\theta))^{3/2}} \, d\theta = \int \frac{\sec^2(\theta)}{\sec^3(\theta)} \, d\theta = \int \cos(\theta) \, d\theta \]

3. **Integrate \( \cos(\theta) \)**:
\[ f(x) = \sin(\theta) + C \]

4. **Convert back to \( x \)**:
Since \( \theta = \tan^{-1}(x) \), we have:
\[ \sin(\theta) = \frac{x}{\sqrt{1+x^2}} \]
Therefore,
\[ f(x) = \frac{x}{\sqrt{1+x^2}} + C \]

5. **Use the initial condition \( f(0) = 0 \)**:
\[ f(0) = \frac{0}{\sqrt{1+0^2}} + C = 0 \implies C = 0 \]
Thus, we have:
\[ f(x) = \frac{x}{\sqrt{1+x^2}} \]

6. **Evaluate \( f(1) \)**:
\[ f(1) = \frac{1}{\sqrt{1+1^2}} = \frac{1}{\sqrt{2}} \]

### Final Answer:
\[ f(1) = \frac{1}{\sqrt{2}} \]