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

Answer

To evaluate the integral

\[ \int \frac{1 + 2x^2}{x^2(1 + x^2)} \, dx, \]

we can follow these steps:

### Step 1: Rewrite the integrand
We can express the integrand as:

\[ \frac{1 + 2x^2}{x^2(1 + x^2)} = \frac{1 + x^2 + x^2}{x^2(1 + x^2)} = \frac{1 + x^2}{x^2(1 + x^2)} + \frac{x^2}{x^2(1 + x^2)}. \]

### Step 2: Split the integral
Now, we can split the integral into two parts:

\[ \int \left( \frac{1 + x^2}{x^2(1 + x^2)} + \frac{x^2}{x^2(1 + x^2)} \right) \, dx = \int \frac{1}{x^2} \, dx + \int \frac{1}{1 + x^2} \, dx. \]

### Step 3: Evaluate each integral
1. **First Integral**:

\[ \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = -\frac{1}{x} + C_1. \]

2. **Second Integral**:

\[ \int \frac{1}{1 + x^2} \, dx = \tan^{-1}(x) + C_2. \]

### Step 4: Combine the results
Now we can combine the results of both integrals:

\[ \int \frac{1 + 2x^2}{x^2(1 + x^2)} \, dx = -\frac{1}{x} + \tan^{-1}(x) + C, \]

where \( C = C_1 + C_2 \) is the constant of integration.

### Final Answer
Thus, the final answer is:

\[ -\frac{1}{x} + \tan^{-1}(x) + C. \]

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