Đáp án:
$I=\ln \dfrac{3}{2}-\dfrac{1}{2}\ln 2.$
Giải thích các bước giải:
$I=\displaystyle\int\limits^1_0 \dfrac{x^3}{x^4+3x^2+2} \, dx\\ t=x^2\Rightarrow dt=2xdx \Rightarrow dx=\dfrac{dt}{2x}\\ \begin{array}{|c|c|}\hline x&0&1\\\hline t&0&1 \\ \hline\end{array}\\ I=\displaystyle\int\limits^1_0 \dfrac{t.x}{t^2+3t+2} \, \dfrac{dt}{2x}\\ =\dfrac{1}{2}\displaystyle\int\limits^1_0 \dfrac{t}{t^2+3t+2} \, dt\\ =\dfrac{1}{2}\displaystyle\int\limits^1_0 \dfrac{t}{(t+1)(t+2)} \, dt\\ \dfrac{t}{(t+1)(t+2)}=\dfrac{A}{t+1}+\dfrac{B}{t+2}\\ \Leftrightarrow \dfrac{t}{(t+1)(t+2)}=\dfrac{A(t+2)+B(t+1)}{(t+1)(t+2)}\\ \Leftrightarrow \dfrac{t}{(t+1)(t+2)}=\dfrac{(A+B)t+2A+B}{(t+1)(t+2)}\\ \text{Đồng nhất hệ số }\Rightarrow \left\{\begin{array}{l} A+B=1 \\2A+B=0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} A=-1 \\B=2 \end{array} \right.\\ \Rightarrow \dfrac{t}{(t+1)(t+2)}=-\dfrac{1}{t+1}+\dfrac{2}{t+2}\\ I=\dfrac{1}{2}\displaystyle\int\limits^1_0 \left(\dfrac{2}{t+2}-\dfrac{1}{t+1}\right) \, dt\\ =\dfrac{1}{2}\left(2\ln |t+2|-\ln |t+1|\right)\Bigg\vert^1_0\\ =\dfrac{1}{2}\left(2\ln 3-\ln 2-2\ln 2-\ln 1\right)\\ =\dfrac{1}{2}\left(2\ln \dfrac{3}{2}-\ln 2\right)\\ =\ln \dfrac{3}{2}-\dfrac{1}{2}\ln 2.$
