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We are given the function \( f(x) = \frac{-2x}{(1 + x^2)^2} \), and its series expansion is written as:

\[
f(x) = \sum_{n=0}^{\infty} (-1)^n 2n x^{2n - 1}
\]

The problem asks us to integrate \( f(x) \) term-by-term from 0 to \( x \) to find the series expansion for the indefinite integral of \( f \).

We need to compute the following integral:

\[
\int_0^x f(t) \, dt = \int_0^x \sum_{n=0}^{\infty} (-1)^n 2n t^{2n - 1} \, dt
\]

### Step-by-Step Solution:

We can integrate the series term-by-term. For each term of the series \( (-1)^n 2n t^{2n - 1} \), the integral becomes:

\[
\int_0^x (-1)^n 2n t^{2n - 1} \, dt = (-1)^n 2n \cdot \frac{t^{2n}}{2n} \bigg|_0^x = (-1)^n x^{2n}
\]

Thus, the series expansion for the integral becomes:

\[
\int_0^x f(t) \, dt = \sum_{n=1}^{\infty} (-1)^n x^{2n}
\]

So, the final expression is:

\[
\int_0^x f(t) \, dt = \sum_{n=1}^{\infty} (-1)^n x^{2n}
\]

Would you like further details on how this integral was computed, or do you have any questions?

### Related Questions:
1. How do you derive the series expansion of a function from scratch?
2. What is the radius of convergence of the series?
3. How does term-by-term integration of a series work, and what are the conditions for it?
4. Can the process be applied to more complicated series or integrals?
5. What are common techniques for integrating power series?

#### Tip: When working with series, always check for convergence before performing operations like integration or differentiation term by term.

Integrate the given series expansion of f term-by-term from zero to x to obtain the corresponding series expansion for the indefinite integral of f.

Learn how to integrate the series expansion of f(x) = -2x/(1 + x^2)^2 term-by-term from zero to x, with detailed steps to obtain the corresponding series expansion for the indefinite integral. This involves power series expansions, term-by-term integration, and the fundamental theorem of calculus.

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