`int_0^oo(pi/(1+pi^2x^2)-1/(1+x^2))logx dx` is equal to (a) `-pi/2logpi` (b) 0 (c) `pi/2log2` (d) none of these

Options

A. `-(pi)/2 In pi`
B. `0`
C. `(pi)/2 In 2`
D. none of these

Explanation

To solve the integral

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

we will break it down into manageable steps.

### Step 1: Rewrite the Integral
We can express the integral \( I \) as:

\[ I = \int_0^\infty \frac{\pi \log x}{1 + \pi^2 x^2} \, dx - \int_0^\infty \frac{\log x}{1 + x^2} \, dx. \]

Let’s denote these two integrals as:

\[ I_1 = \int_0^\infty \frac{\pi \log x}{1 + \pi^2 x^2} \, dx, \]
\[ I_2 = \int_0^\infty \frac{\log x}{1 + x^2} \, dx. \]

Thus, we have:

\[ I = I_1 - I_2. \]

### Step 2: Change of Variable in \( I_1 \)
For \( I_1 \), we will perform the substitution \( y = \pi x \). Then, \( dy = \pi \, dx \) or \( dx = \frac{dy}{\pi} \). The limits remain the same (0 to ∞).

Thus,

\[ I_1 = \int_0^\infty \frac{\pi \log\left(\frac{y}{\pi}\right)}{1 + y^2} \cdot \frac{dy}{\pi} = \int_0^\infty \frac{\log y - \log \pi}{1 + y^2} \, dy. \]

This can be split into two integrals:

\[ I_1 = \int_0^\infty \frac{\log y}{1 + y^2} \, dy - \log \pi \int_0^\infty \frac{1}{1 + y^2} \, dy. \]

### Step 3: Evaluate the Integrals
The integral \( \int_0^\infty \frac{1}{1 + y^2} \, dy \) evaluates to \( \frac{\pi}{2} \).

Thus,

\[ I_1 = \int_0^\infty \frac{\log y}{1 + y^2} \, dy - \frac{\pi}{2} \log \pi. \]

### Step 4: Evaluate \( I_2 \)
The integral \( I_2 \) can be evaluated similarly:

\[ I_2 = \int_0^\infty \frac{\log x}{1 + x^2} \, dx. \]

Notice that \( I_2 \) is equal to \( I_1 \) because both integrals are symmetric in form.

### Step 5: Combine the Results
Now substituting back into our expression for \( I \):

\[ I = \left( \int_0^\infty \frac{\log y}{1 + y^2} \, dy - \frac{\pi}{2} \log \pi \right) - \int_0^\infty \frac{\log x}{1 + x^2} \, dx. \]

Since \( I_1 = I_2 \):

\[ I = 0 - \frac{\pi}{2} \log \pi. \]

### Final Result
Thus, we conclude that:

\[ I = -\frac{\pi}{2} \log \pi. \]

### Answer
The answer is (a) \(-\frac{\pi}{2} \log \pi\).

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