Answer
$x-\frac{x^2}{2}+\frac{x^3}{3}+...\frac{(-1)^nx^{n+1}}{n+1}+...$
Work Step by Step
We are given $f(x)=\ln(1+4x)=\ln(\frac{1}{4}+x)$
for $r$ in $(-\frac{1}{4},\frac{1}{4}]$
The Taylor series for $f(x)=\ln(1+4x)$ is
$f(x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+...+\frac{(-1)^nx^{n+1}}{n+1}+...$
$=x-\frac{x^2}{2}+\frac{x^3}{3}+...\frac{(-1)^nx^{n+1}}{n+1}+...$
