`intlog (1+x^(2))dx.`

Answer

To solve the integral \( \int \log(1 + x^2) \, dx \), we will use the method of integration by parts. Let's go through the steps:

### Step 1: Set up the integration by parts
We will use the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Here, we can choose:
- \( u = \log(1 + x^2) \)
- \( dv = dx \)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{d}{dx} \log(1 + x^2) \, dx = \frac{1}{1 + x^2} \cdot 2x \, dx = \frac{2x}{1 + x^2} \, dx \]
- Integrate \( dv \):
\[ v = \int dx = x \]

### Step 3: Apply the integration by parts formula
Substituting \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula:
\[ \int \log(1 + x^2) \, dx = x \log(1 + x^2) - \int x \cdot \frac{2x}{1 + x^2} \, dx \]

### Step 4: Simplify the integral
Now, simplify the integral:
\[ \int x \cdot \frac{2x}{1 + x^2} \, dx = \int \frac{2x^2}{1 + x^2} \, dx \]
This can be rewritten as:
\[ \int \left( 2 - \frac{2}{1 + x^2} \right) \, dx \]
This is because:
\[ \frac{2x^2}{1 + x^2} = 2 - \frac{2}{1 + x^2} \]

### Step 5: Integrate the simplified expression
Now we can integrate:
\[ \int 2 \, dx - \int \frac{2}{1 + x^2} \, dx = 2x - 2 \tan^{-1}(x) + C \]

### Step 6: Combine the results
Putting it all together:
\[ \int \log(1 + x^2) \, dx = x \log(1 + x^2) - \left( 2x - 2 \tan^{-1}(x) \right) + C \]
This simplifies to:
\[ \int \log(1 + x^2) \, dx = x \log(1 + x^2) - 2x + 2 \tan^{-1}(x) + C \]

### Final Answer
Thus, the final result is:
\[ \int \log(1 + x^2) \, dx = x \log(1 + x^2) - 2x + 2 \tan^{-1}(x) + C \]