`int(x^(2) +3)/(x^(2)+1)dx`

Answer

To solve the integral \(\int \frac{x^2 + 3}{x^2 + 1} \, dx\), we can break it down into simpler parts. Here’s a step-by-step solution:

### Step 1: Rewrite the integrand
We start with the integral:
\[ \int \frac{x^2 + 3}{x^2 + 1} \, dx \]
We can separate the numerator:
\[ \frac{x^2 + 3}{x^2 + 1} = \frac{x^2 + 1 + 2}{x^2 + 1} = \frac{x^2 + 1}{x^2 + 1} + \frac{2}{x^2 + 1} \]
This simplifies to:
\[ 1 + \frac{2}{x^2 + 1} \]

### Step 2: Split the integral
Now we can split the integral into two parts:
\[ \int \left(1 + \frac{2}{x^2 + 1}\right) \, dx = \int 1 \, dx + \int \frac{2}{x^2 + 1} \, dx \]

### Step 3: Integrate each part
1. The integral of 1:
\[ \int 1 \, dx = x \]

2. The integral of \(\frac{2}{x^2 + 1}\):
\[ \int \frac{2}{x^2 + 1} \, dx = 2 \int \frac{1}{x^2 + 1} \, dx \]
We know that:
\[ \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x) \]
Therefore:
\[ 2 \int \frac{1}{x^2 + 1} \, dx = 2 \tan^{-1}(x) \]

### Step 4: Combine the results
Now we combine the results from both integrals:
\[ \int \frac{x^2 + 3}{x^2 + 1} \, dx = x + 2 \tan^{-1}(x) + C \]
where \(C\) is the constant of integration.

### Final Answer
Thus, the final answer is:
\[ \int \frac{x^2 + 3}{x^2 + 1} \, dx = x + 2 \tan^{-1}(x) + C \]

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