Power Series

Remember that the geometric series is $\frac{1}{1-r}=\sum_{n=0}^{\infty}r^n$ for $|r|< 1$

We need to fix up what we have so here we go...

let $g(x)=ln(x+6)$ so then $g'(x)=\frac{1}{x+6}$

then $g'(x)=\frac{\frac{1}{6}}{1-(\frac{-x}{6})}=\frac{1}{6}\sum_{n=0}^{\infty}\left(\frac{-x}{6}\right)^n=\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{6^{n+1}}$

Now integrating both sides we can get g(x)

$\int g'(x)dx= g(x) =\int \sum_{n=0}^{\infty}\frac{(-1)^nx^n}{6^{n+1}}dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{6^{n+1}(n+1)}$

So

$f(x) =7x \cdot g(x)=7x \cdot \sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{6^{n+1}(n+1)}=7 \cdot \sum_{n=0}^{\infty}\frac{(-1)^nx^{n+2}}{6^{n+1}(n+1)}$