The given problem requires finding the integral:

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

Let me solve this step by step.

Step 1: Expand the numerator

The numerator $(1+x)^2$ expands as: $(1+x)^2 = 1 + 2x + x^2$

So the integral becomes: $\int \frac{1 + 2x + x^2}{x^2+1} \, dx$

Step 2: Split the integral

We split the fraction into separate terms: $\frac{1 + 2x + x^2}{x^2+1} = \frac{1}{x^2+1} + \frac{2x}{x^2+1} + \frac{x^2}{x^2+1}$

Thus, the integral becomes: $\int \frac{1}{x^2+1} \, dx + \int \frac{2x}{x^2+1} \, dx + \int \frac{x^2}{x^2+1} \, dx$

Step 3: Solve each term

(a) First term: $\int \frac{1}{x^2+1} \, dx$

This is a standard integral: $\int \frac{1}{x^2+1} \, dx = \arctan(x)$

(b) Second term: $\int \frac{2x}{x^2+1} \, dx$

Substitute $u = x^2 + 1$, so $du = 2x \, dx$. The integral becomes: $\int \frac{2x}{x^2+1} \, dx = \ln(x^2+1)$

(c) Third term: $\int \frac{x^2}{x^2+1} \, dx$

Write $\frac{x^2}{x^2+1}$ as: $\frac{x^2}{x^2+1} = 1 - \frac{1}{x^2+1}$

So the integral becomes: $\int \frac{x^2}{x^2+1} \, dx = \int 1 \, dx - \int \frac{1}{x^2+1} \, dx$ $\int \frac{x^2}{x^2+1} \, dx = x - \arctan(x)$

Step 4: Combine results

Adding all three parts together: $\int \frac{(1+x)^2}{x^2+1} \, dx = \arctan(x) + \ln(x^2+1) + x - \arctan(x)$

Simplify: $\int \frac{(1+x)^2}{x^2+1} \, dx = x + \ln(x^2+1) + C$

Final Answer:

The correct option is: $\boxed{x + \ln(x^2+1) + C}$

Would you like a detailed explanation of any part? Here are related questions to deepen understanding:

1. What are some common substitutions for rational functions in integration?
2. How can you simplify complex fractions for easier integration?
3. What are the standard integrals for trigonometric and logarithmic functions?
4. How does partial fraction decomposition work for integrals?
5. When do you use substitution versus splitting terms in integrals?

Tip: Always look for ways to simplify the numerator and denominator before choosing a method of integration.