`int_(0)^(oo)((ln(1+x^(2)))/(1+x^(2)))dx`.

Answer

To solve the integral

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

we will use a substitution and properties of definite integrals.

### Step-by-Step Solution:

**Step 1: Substitution**
Let \( x = \tan \theta \). Then, \( dx = \sec^2 \theta \, d\theta \).

**Step 2: Change of Limits**
When \( x = 0 \), \( \theta = 0 \). When \( x \to \infty \), \( \theta \to \frac{\pi}{2} \).

**Step 3: Transform the Integral**
Substituting these into the integral gives:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\ln(1+\tan^2 \theta)}{1+\tan^2 \theta} \sec^2 \theta \, d\theta. \]

Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \), we can simplify:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\ln(\sec^2 \theta)}{\sec^2 \theta} \sec^2 \theta \, d\theta = \int_{0}^{\frac{\pi}{2}} \ln(\sec^2 \theta) \, d\theta. \]

**Step 4: Simplify the Logarithm**
Using the property of logarithms, \( \ln(a^b) = b \ln(a) \):

\[ I = 2 \int_{0}^{\frac{\pi}{2}} \ln(\sec \theta) \, d\theta. \]

**Step 5: Change of Variable**
Let \( J = \int_{0}^{\frac{\pi}{2}} \ln(\cos \theta) \, d\theta \). We will relate \( I \) to \( J \).

**Step 6: Use Symmetry Property**
Using the property of definite integrals:

\[ J = \int_{0}^{\frac{\pi}{2}} \ln(\sin \theta) \, d\theta, \]

we can write:

\[ 2J = \int_{0}^{\frac{\pi}{2}} \ln(\sin \theta) \, d\theta + \int_{0}^{\frac{\pi}{2}} \ln(\cos \theta) \, d\theta = \int_{0}^{\frac{\pi}{2}} \ln(\sin \theta \cos \theta) \, d\theta. \]

**Step 7: Use the Double Angle Identity**
Using \( \sin(2\theta) = 2 \sin \theta \cos \theta \):

\[ 2J = \int_{0}^{\frac{\pi}{2}} \ln\left(\frac{1}{2} \sin(2\theta)\right) \, d\theta = \int_{0}^{\frac{\pi}{2}} \ln(\sin(2\theta)) \, d\theta - \int_{0}^{\frac{\pi}{2}} \ln(2) \, d\theta. \]

**Step 8: Change of Variable in the Integral**
Let \( u = 2\theta \), then \( d\theta = \frac{1}{2} du \):

\[ 2J = \frac{1}{2} \int_{0}^{\pi} \ln(\sin u) \, du - \frac{\pi}{2} \ln(2). \]

**Step 9: Evaluate the Integral**
The integral \( \int_{0}^{\pi} \ln(\sin u) \, du = -\pi \ln(2) \). Thus:

\[ 2J = \frac{1}{2} (-\pi \ln(2)) - \frac{\pi}{2} \ln(2) = -\pi \ln(2). \]

**Step 10: Relate Back to \( I \)**
Thus, \( J = -\frac{\pi}{2} \ln(2) \) and substituting back into \( I \):

\[ I = 2J = -\pi \ln(2). \]

### Final Answer:
\[ I = -\pi \ln(2). \]