If $$0 < x < 1$$, then $$\frac{3}{2}x^2 + \frac{5}{3}x^3 + \frac{7}{4}x^4 + \ldots$$, is equal to

Let us denote the required infinite series by $$S$$.

$$S \;=\; \frac{3}{2}x^{2} \;+\; \frac{5}{3}x^{3} \;+\; \frac{7}{4}x^{4} \;+\;\ldots \qquad\text{for} \;0<x<1$$

We first express the general term. For the term containing $$x^{m}$$ (where $$m=2,3,4,\ldots$$) we notice

$$\frac{3}{2},\ \frac{5}{3},\ \frac{7}{4},\ldots$$

Each numerator increases by 2 while each denominator is the exponent itself. Hence for the exponent $$m$$, the coefficient is

$$\frac{2m-1}{m} \;=\; 2 \;-\;\frac{1}{m}.$$ So the series can be written as

$$S \;=\; \sum_{m=2}^{\infty}\left(2-\frac{1}{m}\right)x^{m}.$$

We now split this sum into two simpler sums:

$$S \;=\; 2\sum_{m=2}^{\infty}x^{m}\;-\;\sum_{m=2}^{\infty}\frac{x^{m}}{m}.$$

We evaluate each series separately.

First sum (geometric series): The well-known formula $$\sum_{k=0}^{\infty}x^{k}=\frac{1}{1-x},\quad |x|<1$$ gives us $$\sum_{m=2}^{\infty}x^{m} =\frac{1}{1-x}-\bigl(1+x\bigr) =\frac{1-x^{0}-x^{1}}{1-x} =\frac{x^{2}}{1-x}.$$ Hence $$2\sum_{m=2}^{\infty}x^{m}=2\cdot\frac{x^{2}}{1-x}.$$

Second sum (logarithmic series): The power-series expansion of the natural logarithm is $$-\ln(1-x)=\sum_{k=1}^{\infty}\frac{x^{k}}{k},\quad |x|<1.$$ Therefore $$\sum_{m=2}^{\infty}\frac{x^{m}}{m} =\left(-\ln(1-x)\right)-\frac{x^{1}}{1} =-\ln(1-x)-x.$$

Putting the two evaluated parts back into the expression for $$S$$ we obtain

$$\begin{aligned} S &=2\cdot\frac{x^{2}}{1-x}-\Bigl[-\ln(1-x)-x\Bigr] \\ &=\frac{2x^{2}}{1-x}+\ln(1-x)+x. \end{aligned}$$

To combine the rational terms, we write $$x$$ with the same denominator:

$$x=\frac{x(1-x)}{1-x}.$$

Hence

$$\begin{aligned} \frac{2x^{2}}{1-x}+x &=\frac{2x^{2}}{1-x}+\frac{x(1-x)}{1-x} \\ &=\frac{2x^{2}+x(1-x)}{1-x} \\ &=\frac{2x^{2}+x-x^{2}}{1-x} \\ &=\frac{x^{2}+x}{1-x} \\ &=\frac{x(x+1)}{1-x}. \end{aligned}$$

Therefore the complete value of the series is

$$S=\frac{x(x+1)}{1-x}+\ln(1-x).$$

The given options use $$\log_e$$ for the natural logarithm, so we rewrite $$\ln(1-x)$$ as $$\log_e(1-x)$$ and match with the choices.

Option A is $$x\left(\frac{x+1}{1-x}\right)+\log_e(1-x),$$ which is exactly what we have obtained.

Hence, the correct answer is Option A.