Chọn phương án A.

Đặt \(\begin{cases}
u&=\ln(2x+1)\\
v'&=1
\end{cases}\Rightarrow\begin{cases}
u'&=\dfrac{2}{2x+1}\\
v&=x.
\end{cases}\)

Khi đó ta có $$\begin{aligned}
\displaystyle\int\limits_{0}^{1}\ln(2x+1)\mathrm{\,d}x&=x\ln(2x+1)\bigg|_0^1-\displaystyle\int\limits_{0}^{1}\dfrac{2x}{2x+1}\mathrm{\,d}x\\
&=\ln3-\displaystyle\int\limits_{0}^{1}\dfrac{(2x+1)-1}{2x+1}\mathrm{\,d}x\\
&=\ln3-\displaystyle\int\limits_{0}^{1}\left(1-\dfrac{1}{2x+1}\right)\mathrm{\,d}x\\
&=\ln3-\left(x-\dfrac{\ln|2x+1|}{2}\right)\bigg|_0^1\\
&=\dfrac{3}{2}\ln3-1.
\end{aligned}$$
Vậy \(\begin{cases}
a=\dfrac{3}{2}\\ b=-1
\end{cases}\Rightarrow ab^3=-\dfrac{3}{2}\).