11. $[-/ 2$ Points $]$
DETAILS SCALCET8 11.9.027.
Evaluate the indefinite integral as a power series.
\[
\begin{array}{c}
\int x^{5} \ln (1+x) d x \\
f(x)=C+\sum_{n=1}^{\infty}(\square)
\end{array}
\]
What is the radius of convergence $R$ ?
\[
R=
\]
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Step 1 ：First, we need to find the power series representation of \( \ln(1+x) \). We know that \( \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} \) for \( |x| < 1 \).

Step 2 ：Next, we multiply \( x^5 \) to the power series, which gives us \( x^5 \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{n+5}}{n} \).

Step 3 ：Now, we can integrate term by term to find the indefinite integral. The integral of \( x^{n+5} \) is \( \frac{x^{n+6}}{n+6} \), so the integral of \( x^5 \ln(1+x) \) is \( \int x^5 \ln(1+x) dx = C + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{n+6}}{n(n+6)} \), where C is the constant of integration.

Step 4 ：The radius of convergence R is the same as the original series, which is 1. So, \( R = 1 \).