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$\textbf{Cách 1}$
Áp dụng tích phân từng phần
Đặt $\begin{cases}u=\ln(x+1) \\ dv=(x^{2}+4x+2)dx \end{cases} \Leftrightarrow \begin{cases}du=\dfrac{1}{x+1}dx \\ v=\dfrac{1}{3}(x^3+6x^2 +6x+1)=\dfrac{1}{3}(x+1)(x^2+5x+1)\end{cases}$
Suy ra
$I=uv|_{0}^{1} - \int\limits_{0}^{1}vdu=\left[ {\dfrac{1}{3}(x+1)(x^2+5x+1)\ln(x+1)} \right]_{0}^{1}-\int\limits_{0}^{1}\dfrac{1}{3}(x^2+5x+1)dx$
$=\left[ {\dfrac{1}{3}(x+1)(x^2+5x+1)\ln(x+1)+\dfrac{1}{9}x^3+\dfrac{5}{6}x^2+\dfrac{1}{3}x)} \right]_{0}^{1}$
$=\boxed{-\dfrac{23}{18}+\dfrac{2\ln 2}{3}+\ln 16}$
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