Show that $\ln\left(1+3x + 2x^2\right) = 3x - \frac{5x^2}{2} + \frac{9x^3}{3} -\cdots + \left(-1\right)^{n-1}\frac{2^n+1}{n}x^n+\cdots$ Show that $$\ln\left(1+3x + 2x^2\right) = 3x - \frac{5x^2}{2} + \frac{9x^3}{3} -\cdots + \left(-1\right)^{n-1}\frac{2^n+1}{n}x^n+\cdots$$
I know that $$\ln(1+x) = \sum\limits_{n=0}^{\infty}\frac{\left(-1\right)^nx^{n+1}}{n+1}$$
My first thought was to use a series expansion, $$\ln\left(1+3x+2x^2\right) = \sum\limits_{n=0}^{\infty}\frac{\left(-1\right)^n(3x+2x^2)^{n+1}}{n+1}$$ and compare this sum with the sum on the right hand side. 
$$\sum\limits_{n=0}^{\infty}\frac{\left(-1\right)^n\left(3x+2x^2\right)^{n+1}}{n+1} = \sum\limits_{n=0}^{\infty}\frac{\left(-1\right)^{n-1}\left(2^n+1\right)x^{n}}{n}$$
From here, I would try and manipulate the expression within the sum to give the expression on the right hand side. 
I know I can turn the sum on the left into
$$\sum\limits_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}\left(3x+2x^2\right)^{n}}{n}$$
which is slightly closer to the sum on the right. However, when comparing both sides, the expression I get seems quite incorrect since the highest degree on the left side is $x^{2n}$.
$$\left(3x+2x^2\right)^n = \left(2^n+1\right)x^n$$
Where have I gone wrong? How I can better approach this problem?

Hint: $1+3x+2x^2=(1+x)(1+2x)$, so you just add two Taylor series.
