`int_(0)^(oo)(x logx)/((1+x^(2))^(2)) dx=`

Options

A. 7
B. 0
C. `5 log 13`
D. `2 log5`

Answer

To solve the integral

\[ I = \int_{0}^{\infty} \frac{x \log x}{(1 + x^2)^2} \, dx, \]

we will use the substitution \( x = \frac{1}{t} \). This substitution will help us evaluate the integral over the range from \( 0 \) to \( \infty \).

### Step 1: Substitution
Let \( x = \frac{1}{t} \). Then, we have:
\[ dx = -\frac{1}{t^2} dt. \]
When \( x \to 0 \), \( t \to \infty \) and when \( x \to \infty \), \( t \to 0 \). Therefore, the limits of integration change accordingly.

Substituting into the integral, we get:
\[ I = \int_{\infty}^{0} \frac{\frac{1}{t} \log \frac{1}{t}}{(1 + \left(\frac{1}{t}\right)^2)^2} \left(-\frac{1}{t^2}\right) dt. \]

### Step 2: Simplifying the Integral
Now, we simplify the expression:
\[ \log \frac{1}{t} = -\log t, \]
and
\[ 1 + \left(\frac{1}{t}\right)^2 = 1 + \frac{1}{t^2} = \frac{t^2 + 1}{t^2}. \]
Thus,
\[ (1 + \left(\frac{1}{t}\right)^2)^2 = \left(\frac{t^2 + 1}{t^2}\right)^2 = \frac{(t^2 + 1)^2}{t^4}. \]

Substituting these into the integral gives:
\[ I = \int_{\infty}^{0} \frac{\frac{1}{t} (-\log t)}{\frac{(t^2 + 1)^2}{t^4}} \left(-\frac{1}{t^2}\right) dt. \]
This simplifies to:
\[ I = \int_{0}^{\infty} \frac{\log t}{(t^2 + 1)^2} dt. \]

### Step 3: Combining the Integrals
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{\infty} \frac{x \log x}{(1 + x^2)^2} \, dx \)
2. \( I = \int_{0}^{\infty} \frac{\log t}{(t^2 + 1)^2} dt \)

Since both integrals are equal, we can add them:
\[ 2I = \int_{0}^{\infty} \left( \frac{x \log x}{(1 + x^2)^2} + \frac{\log x}{(x^2 + 1)^2} \right) dx. \]

### Step 4: Evaluating the Integral
Notice that both integrals are symmetric. Therefore, we can conclude:
\[ I = 0. \]

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