We are tasked with evaluating the integral:

\[
I = \int_1^2 \frac{x^2}{(x^3 + 1)^2} \, dx
\]

To solve this, we can try a substitution method. Let's begin by making the substitution:

\[
u = x^3 + 1
\]

This substitution simplifies the denominator. Now, let's differentiate \( u \) with respect to \( x \):

\[
\frac{du}{dx} = 3x^2 \quad \Rightarrow \quad du = 3x^2 \, dx
\]

Thus, \( x^2 \, dx = \frac{du}{3} \).

Now, let's change the limits of integration. When \( x = 1 \), \( u = 1^3 + 1 = 2 \), and when \( x = 2 \), \( u = 2^3 + 1 = 9 \). 

The integral now becomes:

\[
I = \int_2^9 \frac{1}{3u^2} \, du
\]

This simplifies to:

\[
I = \frac{1}{3} \int_2^9 u^{-2} \, du
\]

The integral of \( u^{-2} \) is:

\[
\int u^{-2} \, du = -u^{-1}
\]

So, the integral becomes:

\[
I = \frac{1}{3} \left[ -\frac{1}{u} \right]_2^9
\]

Now, evaluating at the limits:

\[
I = \frac{1}{3} \left( -\frac{1}{9} + \frac{1}{2} \right)
\]

Simplifying:

\[
I = \frac{1}{3} \left( \frac{1}{2} - \frac{1}{9} \right)
= \frac{1}{3} \left( \frac{9 - 2}{18} \right)
= \frac{1}{3} \times \frac{7}{18}
= \frac{7}{54}
\]

Thus, the value of the integral is:

\[
I = \frac{7}{54}
\]

Would you like further clarification or details? Here are 5 related questions you might explore:

1. How does substitution simplify complex integrals?
2. What are the conditions for using substitution in integrals?
3. How do you change the limits of integration after substitution?
4. Why does the power rule for integration apply to functions like \( u^{-2} \)?
5. Can this method be applied to other similar integrals involving polynomials?

**Tip**: Always double-check the limits of integration after substitution to avoid errors.

Evaluate the integral: \int_1^2 \frac{x^2}{(x^3 + 1)^2} \, dx

Explore the step-by-step solution to evaluate the definite integral \int_1^2 \frac{x^2}{(x^3 + 1)^2} dx using the substitution method. This solution covers key calculus concepts including substitution, changing limits, and power rule. Ideal for college-level calculus students.

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