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

Answer

To solve the integral

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

we will use a substitution method.

### Step 1: Substitution
Let \( t = \tan^{-1} x \). Then, we differentiate to find \( dt \):
\[ dt = \frac{1}{1 + x^2} \, dx \quad \Rightarrow \quad dx = (1 + x^2) \, dt \]

### Step 2: Change of Limits
When \( x = 0 \):
\[ t = \tan^{-1}(0) = 0 \]
When \( x \to \infty \):
\[ t = \tan^{-1}(\infty) = \frac{\pi}{2} \]

### Step 3: Rewrite the Integral
Substituting \( x = \tan t \) into the integral, we have:
\[ I = \int_{0}^{\frac{\pi}{2}} t \, dt \]

### Step 4: Evaluate the Integral
Now we can evaluate the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{(\frac{\pi}{2})^2}{2} - \frac{0^2}{2} = \frac{\pi^2}{8} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{\pi^2}{8} \]

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