We have,
$\begin{aligned} & \int x[\log (1+x)]^3 d x=\frac{(1+x)^2}{16} f(x)+(1+x) g(x) \\ & \text { Let } I=\int x[\log (1+x)]^3 d x \\ & \text { put } \log (1+x)=t \Rightarrow 1+x=e^t \Rightarrow d x=e^t d t \\ & \therefore \quad I=\int\left(e^t-1\right) t^3 e^t d t \Rightarrow I=\int\left(e^{2 t} t^3-e^t t^3\right) d t \\ & I=e^{2 t}\left(\frac{t^3}{2}-\frac{3 t^2}{4}+\frac{6 t}{8}-\frac{6}{16}\right) \\ & -e^t\left(t^3-3 t^2+6 t-6\right)+c \\ & I=(1+x)^2 \\ & {\left[\frac{(\log (1+x))^3}{2}-3 \frac{(\log (1+x))^2}{4}+\frac{6 \log (1+x)}{8}-\frac{6}{16}\right]} \\ & -(1+x)\left[(\log (1+x))^3-3(\log (1+x))^2\right. \\ & +6(\log (1+x)-6] \\ & \because \quad f(x)=8(\log (1+x))^3-12(\log (1+x))^2 \\ & +12 \log (1+x)-6 \\ & \text { and, } \quad g(x)=-(\log (1+x))^3+3(\log (1+x))^2 \\ & -6 \log (1+x)+6 \\ & f(x)+g(x)=7(\log (1+x))^3-9(\log (1+x))^2 \\ & +6 \log (1+x)+C \\ & \end{aligned}$