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.