`int_(0)^(oo)((tan^(-1)x)/(x(1+x^(2))))dx`

Answer

To solve the integral

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

we can use the substitution \( x = \tan \theta \). This gives us \( dx = \sec^2 \theta \, d\theta \) and transforms the limits as follows: when \( x = 0 \), \( \theta = 0 \); when \( x \to \infty \), \( \theta \to \frac{\pi}{2} \).

Now substituting these into the integral:

1. **Substituting \( x = \tan \theta \)**:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\tan^{-1}(\tan \theta)}{\tan \theta (1 + \tan^2 \theta)} \sec^2 \theta \, d\theta. \]

2. **Simplifying \( \tan^{-1}(\tan \theta) \)**:
Since \( \tan^{-1}(\tan \theta) = \theta \) for \( \theta \in [0, \frac{\pi}{2}] \), we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\tan \theta (1 + \tan^2 \theta)} \sec^2 \theta \, d\theta. \]

3. **Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \)**:
Thus, the integral simplifies to:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\theta \sec^2 \theta}{\tan \theta \sec^2 \theta} \, d\theta = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\tan \theta} \, d\theta. \]

4. **Expressing \( \frac{1}{\tan \theta} \)**:
We know that \( \frac{1}{\tan \theta} = \cot \theta \):
\[ I = \int_{0}^{\frac{\pi}{2}} \theta \cot \theta \, d\theta. \]

5. **Applying Integration by Parts**:
Let \( u = \theta \) and \( dv = \cot \theta \, d\theta \). Then \( du = d\theta \) and \( v = \ln(\sin \theta) \):
\[ I = \left[ \theta \ln(\sin \theta) \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \ln(\sin \theta) \, d\theta. \]

6. **Evaluating the boundary term**:
At \( \theta = \frac{\pi}{2} \), \( \sin(\frac{\pi}{2}) = 1 \) and \( \ln(1) = 0 \). At \( \theta = 0 \), \( \theta \ln(\sin \theta) \to 0 \) (using L'Hôpital's Rule). Thus, the boundary term evaluates to 0.

7. **Final integral**:
Therefore, we have:
\[ I = -\int_{0}^{\frac{\pi}{2}} \ln(\sin \theta) \, d\theta. \]

8. **Using known results**:
It is known that:
\[ \int_{0}^{\frac{\pi}{2}} \ln(\sin \theta) \, d\theta = -\frac{\pi}{2} \ln(2). \]
Thus, we find:
\[ I = -\left(-\frac{\pi}{2} \ln(2)\right) = \frac{\pi}{2} \ln(2). \]

So, the final answer is:

\[ \int_{0}^{\infty} \frac{\tan^{-1} x}{x(1+x^2)} \, dx = \frac{\pi}{2} \ln(2). \]