Find the value of `int_(0)^(oo)(xlogx)/((1+x^(2))^(2))dx`.

Answer

To solve the integral \( I = \int_0^{\infty} \frac{x \log x}{(1+x^2)^2} \, dx \), we will use a substitution and properties of logarithms to simplify the expression.

### Step 1: Substitution
Let \( x = \frac{1}{t} \). Then, as \( x \) goes from \( 0 \) to \( \infty \), \( t \) goes from \( \infty \) to \( 0 \). The differential \( dx \) can be computed as follows:
\[ dx = -\frac{1}{t^2} \, dt \]

### Step 2: Change the limits and rewrite the integral
Substituting \( x \) and \( dx \) into the integral, we have:
\[ I = \int_{\infty}^{0} \frac{\frac{1}{t} \log \left( \frac{1}{t} \right)}{\left( 1 + \left( \frac{1}{t} \right)^2 \right)^2} \left( -\frac{1}{t^2} \right) dt \]
This simplifies to:
\[ I = \int_{0}^{\infty} \frac{\frac{1}{t} (-\log t)}{\left( 1 + \frac{1}{t^2} \right)^2} \frac{1}{t^2} dt \]

### Step 3: Simplify the expression
Now simplifying the denominator:
\[ 1 + \frac{1}{t^2} = \frac{t^2 + 1}{t^2} \]
Thus,
\[ \left( 1 + \frac{1}{t^2} \right)^2 = \left( \frac{t^2 + 1}{t^2} \right)^2 = \frac{(t^2 + 1)^2}{t^4} \]
Now substituting this back into the integral:
\[ I = \int_{0}^{\infty} \frac{-\log t}{t^3} \cdot \frac{t^4}{(t^2 + 1)^2} dt = \int_{0}^{\infty} \frac{-t \log t}{(t^2 + 1)^2} dt \]

### Step 4: Combine the integrals
Now we have:
\[ I = -\int_{0}^{\infty} \frac{t \log t}{(t^2 + 1)^2} dt \]
We can denote this integral as \( J \):
\[ J = \int_{0}^{\infty} \frac{t \log t}{(t^2 + 1)^2} dt \]
Thus, we can express \( I \) as:
\[ I = -J \]

### Step 5: Adding the two integrals
Now, we can observe that:
\[ I + J = 0 \implies 2I = 0 \implies I = 0 \]

### Conclusion
Thus, the value of the integral is:
\[ \int_0^{\infty} \frac{x \log x}{(1+x^2)^2} \, dx = 0 \]