Let `f(x)=int x^2/((1+x^2)(1+sqrt(1+x^2)))dx` and `f(0)=0` then `f(1)` is

Options

A. an increasing function
B. a decreasing function
C. a non-monotonic function
D. can't say anything

Explanation

To solve the problem, we need to evaluate the integral given by the function \( f(x) = \int \frac{x^2}{(1+x^2)(1+\sqrt{1+x^2})} \, dx \) and find \( f(1) \) given that \( f(0) = 0 \).

### Step-by-Step Solution:

1. **Set up the integral**:
\[ f(x) = \int \frac{x^2}{(1+x^2)(1+\sqrt{1+x^2})} \, dx \]

2. **Substitute \( x = \tan \theta \)**:
Using the substitution \( x = \tan \theta \), we have \( dx = \sec^2 \theta \, d\theta \) and \( 1 + x^2 = \sec^2 \theta \). Thus, we can rewrite the integral:
\[ f(x) = \int \frac{\tan^2 \theta}{\sec^2 \theta (1+\sqrt{\sec^2 \theta})} \sec^2 \theta \, d\theta \]
Simplifying gives:
\[ f(x) = \int \frac{\tan^2 \theta}{1+\sqrt{\sec^2 \theta}} \, d\theta \]

3. **Simplify the denominator**:
Note that \( \sqrt{\sec^2 \theta} = \sec \theta \), so:
\[ f(x) = \int \frac{\tan^2 \theta}{1+\sec \theta} \, d\theta \]

4. **Rewrite \( \tan^2 \theta \)**:
Using the identity \( \tan^2 \theta = \sec^2 \theta - 1 \):
\[ f(x) = \int \frac{\sec^2 \theta - 1}{1+\sec \theta} \, d\theta \]

5. **Split the integral**:
This can be split into two separate integrals:
\[ f(x) = \int \frac{\sec^2 \theta}{1+\sec \theta} \, d\theta - \int \frac{1}{1+\sec \theta} \, d\theta \]

6. **Evaluate the first integral**:
The first integral can be evaluated using the substitution \( u = 1 + \sec \theta \), leading to:
\[ \int \frac{\sec^2 \theta}{1+\sec \theta} \, d\theta = \ln |1 + \sec \theta| + C \]

7. **Evaluate the second integral**:
The second integral can be evaluated similarly:
\[ \int \frac{1}{1+\sec \theta} \, d\theta = \ln |1 + \sec \theta| - \theta + C \]

8. **Combine results**:
Combining these results gives:
\[ f(x) = \ln |1 + \sec \theta| - \left( \ln |1 + \sec \theta| - \theta \right) + C \]
Simplifying leads to:
\[ f(x) = -\theta + C \]

9. **Back-substitute for \( \theta \)**:
Since \( \theta = \tan^{-1}(x) \):
\[ f(x) = -\tan^{-1}(x) + C \]

10. **Use the condition \( f(0) = 0 \)**:
Plugging in \( x = 0 \):
\[ f(0) = -\tan^{-1}(0) + C = 0 \implies C = 0 \]
Thus, we have:
\[ f(x) = -\tan^{-1}(x) \]

11. **Find \( f(1) \)**:
Now, we calculate \( f(1) \):
\[ f(1) = -\tan^{-1}(1) = -\frac{\pi}{4} \]

### Final Answer:
\[ f(1) = -\frac{\pi}{4} \]