The given function is:

$f(x) = \frac{9}{1 - x}$

We are tasked with finding the power series of the integral:

$I(x) = \int_0^x f(t) \, dt = \int_0^x \frac{9}{1 - t} \, dt$

Step 1: Power Series Expansion of $f(t) = \frac{9}{1 - t}$

First, we recognize that the function $f(t) = \frac{9}{1 - t}$ can be written as a geometric series expansion:

$\frac{1}{1 - t} = \sum_{n=0}^{\infty} t^n \quad \text{(for } |t| < 1\text{)}$

Thus, we have:

$f(t) = 9 \cdot \frac{1}{1 - t} = 9 \sum_{n=0}^{\infty} t^n = 9 \left( 1 + t + t^2 + t^3 + \dots \right)$

Step 2: Integrating the Series Term-by-Term

Now, we integrate this power series term-by-term from 0 to $x$:

$I(x) = \int_0^x f(t) \, dt = \int_0^x 9 \sum_{n=0}^{\infty} t^n \, dt$

Since the series converges for $|x| < 1$, we can integrate each term:

$I(x) = 9 \sum_{n=0}^{\infty} \int_0^x t^n \, dt = 9 \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$

Step 3: First 4 Non-Zero Terms

Now, we extract the first 4 non-zero terms of this power series:

$I(x) = 9 \left( \frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots \right)$

Thus, the first four terms are:

$I(x) = 9x + \frac{9x^2}{2} + \frac{9x^3}{3} + \frac{9x^4}{4} + \dots$

Step 4: General Term of the Series

The general term of the series is:

$\frac{9x^{n+1}}{n+1}$

Thus, the general term is:

$\frac{9x^{n+1}}{n+1}, \quad n = 0, 1, 2, \dots$

Final Answer

• First 4 non-zero terms:

$9x + \frac{9x^2}{2} + \frac{9x^3}{3} + \frac{9x^4}{4}$

• General term:

$\frac{9x^{n+1}}{n+1}$

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Would you like any further details or explanations on this? Here are some related questions for you:

1. How does the convergence of the series change with different values of $x$?
2. What would the series expansion look like for a different function, such as $\frac{9}{1 + x}$?
3. How does the integral of a geometric series relate to other common power series expansions?
4. How can we use this integral result for applications like solving differential equations?
5. What are some methods for approximating integrals numerically when the series is too complex to sum directly?

Tip: For power series, ensure that the radius of convergence is understood and respected. If you're working with a geometric series, remember that the series only converges when $|x| < 1$.