ĐGNL ĐHQG Hà Nội - Tư duy định lượng - Sử dụng phương pháp nguyên hàm từng phần để tìm nguyên hàm

$$\int udv=uv+\int vdu$$

$$\int udv=uv-\int vdu$$

$$\int udv=\int uv-\int vdu$$

$$\int udv=\int uvdv-\int vdu$$

$\{\begin{array}{c}du=g^{\prime }\left(x\right)dx\\ v=\int h(x)dx\end{array}$

$\{\begin{array}{c}du=g\left(x\right)dx\\ v=\int h(x)dx\end{array}$

$\{\begin{array}{c}du=\int g\left(x\right)dx\\ v=h(x)dx\end{array}$

$\{\begin{array}{c}du=g^{\prime }\left(x\right)dx\\ v=h(x)dx\end{array}$

$$I=\left(x+1\right)f\left(x\right)-2F\left(x\right)+C$$

$$I=F\left(x\right)-\left(x+1\right)f\left(x\right)$$

$$I=\left(x+1\right)f\left(x\right)+C$$

$$I=\left(x+1\right)f\left(x\right)-F\left(x\right)+C$$

$$\int f(x)dx=x^3\ln 3x-\frac{x^3}{3}+C$$

$$\int f(x)dx=\frac{x^3\ln 3x}{3}-\frac{x^3}{9}+C$$

$$\int f(x)dx=\frac{x^3\ln 3x}{3}-\frac{x^3}{3}+C$$

$$\int f(x)dx=\frac{x^3\ln 3x}{3}-\frac{x^3}{27}+C$$

−1

3

11

2

$$a=2$$

$$a=-1$$

$$a=0$$

$$a=1$$

$$F\left(x\right)=\frac{2^x+\ln 2-1}{e^x\left(\ln 2-1\right)}$$

$$F\left(x\right)=\frac{1}{\ln 2-1}{\left(\frac{2}{e}\right)}^x+{\left(\frac{1}{e}\right)}^x-\frac{1}{\ln 2-1}$$

$$F\left(x\right)=\frac{2^x+\ln 2}{e^x\left(\ln 2-1\right)}$$

$$F\left(x\right)={\left(\frac{2}{e}\right)}^x$$

$$\frac{1}{2}\left(\frac{1}{4}\sin 2x-\frac{x}{2}\cos 2x\right)+C$$

$$-\frac{1}{2}\left(\frac{1}{2}\sin 2x-\frac{x}{4}\cos 2x\right)+C$$

$$\frac{1}{2}\left(\frac{1}{2}\sin 2x+\frac{x}{2}\cos 2x\right)+C$$

$$-\frac{1}{2}\left(\frac{1}{2}\sin 2x+\frac{x}{4}\cos 2x\right)+C$$

$$2\left(\sqrt{x}\sin \sqrt{x}-\cos \sqrt{x}\right)+C$$

$$2\left(\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}\right)+C$$

$$\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}+C$$

$$\sqrt{x}\sin \sqrt{x}-\cos \sqrt{x}+C$$

F(x) là hàm chẵn.

F(x) là hàm lẻ.

F(x) là hàm tuần hoàn với chu kì 2π.

F(x) không là hàm chẵn cũng không là hàm lẻ.

$$F\left(\pi \right)=-1$$

$$F\left(\pi \right)=\frac{1}{2}$$

$$F\left(\pi \right)=1$$

$$F\left(\pi \right)=0$$

$$-\frac{1}{2}{x}^2+x\tan x+\ln \left|\cos x\right|+C$$

$$-\frac{1}{2}{x}^2+x\tan x-\ln \left|\cos x\right|+C$$

$$\frac{1}{2}{x}^2+x\tan x-\ln \left|\cos x\right|+C$$

$$\frac{1}{2}{x}^2-x\tan x+\ln \left|\cos x\right|+C$$

$$I=\frac{1}{2}\left(1+\sin 2x\right)\ln \left(1+\sin 2x\right)-\frac{1}{4}\sin 2x+C$$

$$I=\frac{1}{4}\left(1+\sin 2x\right)\ln \left(1+\sin 2x\right)-\frac{1}{2}\sin 2x+C$$

$$I=\frac{1}{4}\left(1+\sin 2x\right)\ln \left(1+\sin 2x\right)-\frac{1}{4}\sin 2x+C$$

$$I=\frac{1}{4}\left(1+\sin 2x\right)\ln \left(1+\sin 2x\right)+\frac{1}{4}\sin 2x+C$$

$$x\ln \left(x+\sqrt{x^2+1}\right)-\sqrt{x^2+1}+C$$

$$\ln \left(x+\sqrt{x^2+1}\right)-\sqrt{x^2+1}+C$$

$$x\ln \left(x+\sqrt{x^2+1}\right)+\sqrt{x^2+1}+C$$

$$\ln \left(x+\sqrt{x^2+1}\right)+\sqrt{x^2+1}+C$$

$$\frac{e^{2x}}{13}\left(2\sin 3x+3\cos 3x\right)+C$$

$$\frac{e^{2x}}{13}\left(3\sin 3x-2\cos 3x\right)+C$$

$$\frac{e^{2x}}{13}\left(2\sin 3x-3\cos 3x\right)+C$$

$$\frac{e^{2x}}{13}\left(3\sin 3x+2\cos 3x\right)+C$$

$$F\left(x\right)=x{e}^x+1-\ln \left|x{e}^x+1\right|+C$$

$$F\left(x\right)=e^x+1-\ln \left|x{e}^x+1\right|+C$$

$$F\left(x\right)=x{e}^x+1-\ln \left|x{e}^{-x}+1\right|+C$$

$$F\left(x\right)=x{e}^x+1+\ln \left|x{e}^x+1\right|+C$$

$$\frac{x}{x^2+1}+C$$

$$\frac{2x}{x^2+1}+C$$

$$\frac{-x}{x^2+1}+C$$

$$\frac{-2x}{x^2+1}+C$$

$$\frac{7}{2}$$

$$\frac{5-e}{2}$$

$$\frac{7-e}{2}$$

$$\frac{5}{2}$$

$$\left(x+1\right)e^x+C$$

$$\left(x+1\right)e^x-x+C$$

$$\left(x+2\right)e^x-x+C$$

$$\left(x+1\right)e^x+x+C$$

$$2020\left(x-2\right)e^x+C$$

$$x{e}^x+C$$

$$2020\left(x+2\right)e^x+C$$

$$\left(x-2\right)e^x+C$$