Integrate the series for $\frac { 1 } { x + 6 }$ to find the power series for the function $f ( x ) = \ln ( x + 6 )$ .
A) $\frac { 1 } { 7 } \sum _ { n = 0 } ^ { \infty } \left( \frac { - 1 } { 7 } \right) ^ { n } \frac { ( x ) ^ { n } } { n + 1 }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( 6 x - 1 ) ^ { n - 1 } } { n - 1 }$
C) $\frac { 1 } { 6 } \sum _ { n = 0 } ^ { \infty } \frac { ( x ) ^ { n + 1 } } { n + 1 }$
D) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( 6 x - 1 ) ^ { n } } { n - 1 }$
E) $\frac { 1 } { 6 } \sum _ { n = 0 } ^ { \infty } \left( \frac { - 1 } { 6 } \right) ^ { n } \frac { ( x ) ^ { n + 1 } } { n + 1 }$

Question

Quizplus · Accepted Answer

The power series for $\frac{1}{x+6}$ is $\sum_{n=0}^{\infty} \left(-\frac{x}{6}\right)^n$. Integrating term by term gives $\ln(x+6) = \int \frac{1}{x+6} \, dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{6^n} \frac{x^{n+1}}{n+1}$, which matches option E.

Integrate the Series For 1x+6\frac { 1 } { x + 6 }x+61​

Integrate the Series For $\frac { 1 } { x + 6 }$