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The integral provided is:

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

To solve this integral, we can use the substitution method. Here's how:

### Step 1: Substitution
Let:
\[
u = x^2 + 1
\]
Then, the derivative of \( u \) with respect to \( x \) is:
\[
\frac{du}{dx} = 2x \quad \Rightarrow \quad du = 2x \, dx
\]

### Step 2: Rewriting the Integral
The integral now becomes:
\[
\int \frac{x^2 \, dx}{(x^2 + 1)^2} = \int \frac{x^2}{u^2} \cdot \frac{du}{2x}
\]

Simplifying, we get:
\[
\frac{1}{2} \int \frac{x \, du}{u^2}
\]

However, we can simplify this further since \( x^2 = u - 1 \) from our substitution \( u = x^2 + 1 \):
\[
\frac{1}{2} \int \frac{u - 1}{u^2} \, du
\]

### Step 3: Break into Two Separate Integrals
Now, break it into two integrals:
\[
\frac{1}{2} \left( \int \frac{du}{u} - \int \frac{du}{u^2} \right)
\]

These are standard integrals:
1. \(\int \frac{du}{u} = \ln|u|\)
2. \(\int \frac{du}{u^2} = -\frac{1}{u}\)

So the expression becomes:
\[
\frac{1}{2} \left( \ln|u| + \frac{1}{u} \right) + C
\]

### Step 4: Back Substitution
Substitute back \( u = x^2 + 1 \):
\[
\frac{1}{2} \left( \ln|x^2 + 1| + \frac{1}{x^2 + 1} \right) + C
\]

Thus, the solution to the integral is:
\[
\int \frac{x^2 \, dx}{(x^2 + 1)^2} = \frac{1}{2} \ln|x^2 + 1| + \frac{1}{2(x^2 + 1)} + C
\]

Would you like more details on this solution, or do you have any further questions?

### Follow-Up Questions
1. What are some other methods that could be used to solve similar integrals?
2. How does the substitution method simplify the process of solving integrals?
3. Can this integral be solved using partial fractions? Why or why not?
4. What are some common integrals that use the logarithm function in their solutions?
5. How does the choice of substitution affect the complexity of the integral solution?

### Tip
Always check if simplifying the integrand before substituting can make the integral easier to solve.

Learn how to solve the integral ∫ x^2 / (x^2 + 1)^2 dx using the substitution method. This problem involves key calculus concepts, including integration by substitution and logarithmic functions. Suitable for advanced high school students and college-level calculus.

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