đề phải cho 1 vecto là cchieeuf cao thì ms tính đl chứ

vd như OA', OB', OC' OD',.. j đấy chứ b

\(\int\limits^2_1\dfrac{lnx}{x^2}.dx\)

Dat \(\left\{{}\begin{matrix}u=lnx\\dv=\dfrac{dx}{x^2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=-\dfrac{1}{x}\end{matrix}\right.\)

\(\Rightarrow\int\limits^2_1\dfrac{lnx}{x^2}dx=lnx.\left(-\dfrac{1}{x}\right)|^2_1+\int\limits^2_1\dfrac{1}{x^2}.dx\)

\(=lnx.\left(-\dfrac{1}{x}\right)|^2_1+\left(-\dfrac{1}{x}\right)|^2_1=\left(-\dfrac{1}{2}\right).ln2+ln1-\dfrac{1}{2}+1\)

\(=\dfrac{1}{2}-\dfrac{1}{2}ln2\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=1\\c=2\end{matrix}\right.\Rightarrow P=2a+3b+c=-1+3+2=4\)

Đặt \(x=4sint\Rightarrow\left\{{}\begin{matrix}dx=4cost.dt\\x=0\Rightarrow t=0\\x=2\sqrt{2}\Rightarrow t=\dfrac{\pi}{4}\end{matrix}\right.\)

\(I=\int\limits^{\dfrac{\pi}{4}}_04.cost.4cost.dt=16\int\limits^{\dfrac{\pi}{4}}_0cos^2tdt=8\int\limits^{\dfrac{\pi}{4}}_0\left(1+cos2t\right)dt\)

\(=8\left(x+\dfrac{1}{2}sin2t\right)|^{\dfrac{\pi}{4}}_0=...\)

\(I=\int e^xcos2xdx\)

Đặt \(\left\{{}\begin{matrix}u=e^x\\dv=cos2xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=e^xdx\\v=\dfrac{1}{2}sin2x\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{2}e^xsin2x-\dfrac{1}{2}\int e^xsin2xdx\)

Xét \(I_1=\int e^xsin2xdx\)

Đặt \(\left\{{}\begin{matrix}u=e^x\\dv=sin2xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=e^xdx\\v=-\dfrac{1}{2}cos2x\end{matrix}\right.\)

\(\Rightarrow I_1=-\dfrac{1}{2}e^xcos2x+\dfrac{1}{2}\int e^xcos2xdx=-\dfrac{1}{2}e^xcos2x+\dfrac{1}{2}I\)

\(\Rightarrow I=\dfrac{1}{2}e^xsin2x-\dfrac{1}{2}\left(-\dfrac{1}{2}e^xcos2x+\dfrac{1}{2}I\right)\)

\(\Rightarrow\dfrac{5}{4}I=\dfrac{1}{2}e^xsin2x+\dfrac{1}{4}e^xcos2x+C\)

\(\Rightarrow I=\dfrac{2}{5}e^xsin2x+\dfrac{1}{5}e^xcos2x+C\)

\(I=\dfrac{1}{2}\int f\left(x^2\right).d\left(x^2\right)=\dfrac{1}{2}x^2\sqrt{\left(x^2\right)^2+1}=\dfrac{1}{2}x^2\sqrt{x^4+1}\)