Đặt \(\left\{{}\begin{matrix}u=ln\left(2x+1\right)\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{2}{2x+1}dx\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{2}x^2.ln\left(2x+1\right)|^3_0-\int\limits^3_0\dfrac{x^2}{2x+1}dx=\dfrac{9}{2}ln7-\dfrac{1}{4}\int\limits^3_0\left(2x-1+\dfrac{1}{2x+1}\right)dx\)
\(=\dfrac{9}{2}ln10-\dfrac{1}{4}\left(x^2-x+\dfrac{1}{2}ln\left(2x+1\right)\right)|^3_0=\dfrac{35}{8}ln7-\dfrac{3}{2}\)
\(\Rightarrow S=7-3+2=6\)
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