The value of the integral `int_0^oo(xlogx)/((1+x^2)^2)dx ` (a) 0 (b) log 7 (c) 5 log 13 (d) none of these

Options

A. `0`
B. `log7`
C. `5 log 13`
D. none of these

Explanation

To solve the integral

\[ I = \int_0^\infty \frac{x \log x}{(1 + x^2)^2} \, dx, \]

we will use the substitution \( x = \frac{1}{t} \).

### Step 1: Substitution
Let \( x = \frac{1}{t} \). Then, the differential \( dx \) becomes:

\[ dx = -\frac{1}{t^2} \, dt. \]

### Step 2: Change of Limits
When \( x \to 0 \), \( t \to \infty \), and when \( x \to \infty \), \( t \to 0 \). Thus, the limits of integration change from \( 0 \) to \( \infty \) to \( \infty \) to \( 0 \).

### Step 3: Rewrite the Integral
Substituting into the integral, we have:

\[ I = \int_\infty^0 \frac{\frac{1}{t} \log \left(\frac{1}{t}\right)}{\left(1 + \left(\frac{1}{t}\right)^2\right)^2} \left(-\frac{1}{t^2}\right) dt. \]

### Step 4: Simplify the Integral
Now, we simplify the expression:

\[ \log\left(\frac{1}{t}\right) = -\log t, \]

and

\[ 1 + \left(\frac{1}{t}\right)^2 = 1 + \frac{1}{t^2} = \frac{t^2 + 1}{t^2}. \]

Thus,

\[ \left(1 + \left(\frac{1}{t}\right)^2\right)^2 = \left(\frac{t^2 + 1}{t^2}\right)^2 = \frac{(t^2 + 1)^2}{t^4}. \]

Substituting these into the integral gives:

\[ I = \int_\infty^0 \frac{\frac{1}{t} (-\log t)}{\frac{(t^2 + 1)^2}{t^4}} \left(-\frac{1}{t^2}\right) dt. \]

### Step 5: Rearranging the Integral
This simplifies to:

\[ I = \int_\infty^0 \frac{\log t}{t} \cdot \frac{t^4}{(t^2 + 1)^2} \cdot \frac{1}{t^2} dt = \int_\infty^0 \frac{t^2 \log t}{(t^2 + 1)^2} dt. \]

### Step 6: Change Limits
Reversing the limits of integration gives:

\[ I = \int_0^\infty \frac{t^2 \log t}{(t^2 + 1)^2} dt. \]

### Step 7: Combine Integrals
Now we have:

\[ I = -I. \]

This implies:

\[ 2I = 0 \implies I = 0. \]

### Conclusion
Thus, the value of the integral is:

\[ \boxed{0}. \]