Answer:
To calculate the integral \(\int \frac{x^3}{(1-x)^2} dx\) using the series method, we can expand the integrand as a power series and then integrate each term term by term. Here's how you can do it:
First, we need to find the power series expansion of \(\frac{1}{(1-x)^2}\). We can use the geometric series formula:
\(\frac{1}{1 - u} = \sum_{n=0}^{\infty} u^n\)
In our case, \(u = x\) and we want the expansion for \(\frac{1}{(1-x)^2}\), so we'll differentiate both sides of the above formula with respect to \(u\) and then multiply by \(-1\) to get the expansion for \(\frac{1}{(1-x)^2}\):
\(\frac{d}{du}\left(\frac{1}{1 - u}\right) = \frac{d}{du}\left(\sum_{n=0}^{\infty} u^n\right)\)
\(-\frac{1}{(1-u)^2} = \sum_{n=0}^{\infty} nu^{n-1}\)
Now, we have the series expansion of \(\frac{1}{(1-x)^2}\):
\(-\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty} nu^{n-1}\)
Next, let's multiply both sides by \(x^3\) to get the series expansion of \(\frac{x^3}{(1-x)^2}\):
\(-x^3\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty} nx^{n+2}\)
Now, we can integrate each term of this series with respect to \(x\) term by term. The result will be a power series, and we can find the antiderivative of each term easily:
\(\int \frac{x^3}{(1-x)^2} dx = \sum_{n=0}^{\infty} \int nx^{n+2} dx\)
Now, integrate each term:
\(\int nx^{n+2} dx = \frac{n}{n+3}x^{n+3} + C\)
So, the antiderivative of \(\frac{x^3}{(1-x)^2}\) is given by:
\(\int \frac{x^3}{(1-x)^2} dx = \sum_{n=0}^{\infty} \frac{n}{n+3}x^{n+3} + C\)
This is the result as a power series. However, keep in mind that this is an infinite series, and finding a closed-form expression for the sum may not always be possible. Depending on the specific range of integration, you might need to truncate the series and work with a finite number of terms for practical calculations.
Explanation: