The value of the integral `int_(0)^(oo) (x log x)/((1+x^(2))^(2)) dx` is

Options

A. 1
B. 0
C. 2
D. None of these

Answer

To solve the integral \( I = \int_{0}^{\infty} \frac{x \log x}{(1+x^2)^2} \, dx \), we can use a substitution and properties of definite integrals. Let's go through the solution step by step.

### Step 1: Substitution
We will use the substitution \( x = \tan \theta \). Then, we have:
\[ dx = \sec^2 \theta \, d\theta \]
The limits of integration change as follows:
- When \( x = 0 \), \( \theta = \tan^{-1}(0) = 0 \).
- When \( x = \infty \), \( \theta = \tan^{-1}(\infty) = \frac{\pi}{2} \).

### Step 2: Change the integral
Substituting \( x = \tan \theta \) into the integral, we get:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\tan \theta \log(\tan \theta)}{(1+\tan^2 \theta)^2} \sec^2 \theta \, d\theta \]
Since \( 1 + \tan^2 \theta = \sec^2 \theta \), we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\tan \theta \log(\tan \theta)}{\sec^4 \theta} \sec^2 \theta \, d\theta = \int_{0}^{\frac{\pi}{2}} \tan \theta \log(\tan \theta) \cos^2 \theta \, d\theta \]

### Step 3: Simplifying the integral
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin \theta \log(\tan \theta)}{\cos \theta} \cos^2 \theta \, d\theta = \int_{0}^{\frac{\pi}{2}} \sin \theta \log(\tan \theta) \cos \theta \, d\theta \]

### Step 4: Using symmetry
We can use the property of definite integrals:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \]
Let \( J = \int_{0}^{\frac{\pi}{2}} \cos \theta \log(\cos \theta) \sin \theta \, d\theta \). Then,
\[ I = \int_{0}^{\frac{\pi}{2}} \sin \theta \log(\tan \theta) \cos \theta \, d\theta \]
and
\[ I = \int_{0}^{\frac{\pi}{2}} \cos \theta \log(\cos \theta) \sin \theta \, d\theta \]

### Step 5: Adding the two integrals
Adding these two equations gives:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \sin \theta \cos \theta (\log(\tan \theta) + \log(\cos \theta)) \, d\theta \]
Using the property \( \log a + \log b = \log(ab) \):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \sin \theta \cos \theta \log(\sin \theta) \, d\theta \]

### Step 6: Evaluating the integral
The integral \( \int_{0}^{\frac{\pi}{2}} \sin \theta \cos \theta \log(\sin \theta) \, d\theta \) can be shown to be zero, leading to:
\[ 2I = 0 \implies I = 0 \]

### Conclusion
Thus, the value of the integral is:
\[ \boxed{0} \]