`int x^2/(1+x^2) dx`

Answer

To solve the integral \(\int \frac{x^2}{1+x^2} \, dx\), we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the integrand \(\frac{x^2}{1+x^2}\) by adding and subtracting 1 in the numerator:
\[ \frac{x^2}{1+x^2} = \frac{x^2 + 1 - 1}{1+x^2} = \frac{x^2 + 1}{1+x^2} - \frac{1}{1+x^2} \]
This gives us:
\[ \int \frac{x^2}{1+x^2} \, dx = \int \left( \frac{x^2 + 1}{1+x^2} - \frac{1}{1+x^2} \right) \, dx \]

### Step 2: Separate the integrals
Now we can separate the integral into two parts:
\[ \int \frac{x^2 + 1}{1+x^2} \, dx - \int \frac{1}{1+x^2} \, dx \]

### Step 3: Simplify the first integral
The first integral simplifies as follows:
\[ \frac{x^2 + 1}{1+x^2} = 1 \]
Thus, we have:
\[ \int \frac{x^2 + 1}{1+x^2} \, dx = \int 1 \, dx = x \]

### Step 4: Evaluate the second integral
The second integral is a standard integral:
\[ \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) \]

### Step 5: Combine the results
Putting it all together, we have:
\[ \int \frac{x^2}{1+x^2} \, dx = x - \tan^{-1}(x) + C \]
where \(C\) is the constant of integration.

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

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