Geometric Power Series Representation of ln(1+2x) at c=0

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I was wondering if someone could check my work:

Find the geometric power series representation of
f(x)=ln(1+2x), c=0

I get \\sum_{n=0}^ \\infty2(-2x)^n+1 on -1/2<x<1/2
 
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f(x) = \ln(1+2x)

\ln(1+2x) = \frac{1}{2} \int \frac{1}{1+2x}

\int \frac{1}{1+2x} = \int \sum_{n=0}^{\infty} (-2x)^{n} = \sum_{n=0}^{\infty} \frac{(-2x)^{n+1}}{n+1}


\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-2x)^{n+1}}{n+1} = \sum_{n=0}^{\infty} \frac{(-2x)^{n+1}}{2n+2}
 
Thanks! I see what I did.
 
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