Reciprocal logarithm integration

$I=\int_{-1}^0 \frac{x^2 + 2x}{\ln(x+1)}dx = \int_{-1}^0 \frac{x^2 + 2x {\color{#D61F06} + 1 - 1}}{\ln(x+1)}dx = \int_{-1}^0 \frac{(x+1)^2 - 1}{\ln(x+1)}dx\\$

suppose that $I(a) = \int_{-1}^0 \frac{(x+1)^a - 1}{\ln(x+1)}dx$

our goal is $I(2)$

$\begin{aligned} I(a) & = \int_{-1}^0 \frac{(x+1)^a - 1}{\ln(x+1)}dx\\ \frac{d}{da} I(a)& = \int_{-1}^0 (x+1)^a \ dx = \left . \frac{(x+1)^{a+1}}{a+1} \right |_{-1}^0 =\frac{1}{a+1} \\ I(a)& = \ln(a+1) + C \\ &{\color{#D61F06} I(0) =\int_{-1}^0 \frac{(x+1)^0 - 1}{\ln(x+1)}dx = 0 } \\ I(0)& = \ln(0+1) + C \Rightarrow C = 0 \\ I(a)& = \ln(a+1) \\ I(2)& = \ln(2+1) =\ln(3) \\ \end{aligned}$