ĐGTD ĐH Bách khoa - Tư duy Toán học - Nguyên hàm (từng phần)

$\int udv=uv+\int vdu$

$\int udv=uv-\int vdu$

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

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

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

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

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

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

a = 2

a = -1

a = 0

a = 1

-1

3

11

2

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

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

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

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

a + b = 2

a + b = 3

a + b = 0

a + b = 1

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.

$\frac{1}{4}{x}^4\ln 3x+C$

$-\frac{1}{4}{x}^4\ln 3x-\frac{1}{16}{x}^4+C$

$-\frac{1}{4}{x}^4\ln 3x+\frac{1}{16}{x}^4+C$

$\frac{1}{4}{x}^4\ln 3x-\frac{1}{16}{x}^4+C$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\frac{7}{2}$

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

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

$\frac{5}{2}$

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

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

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

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

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

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

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

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

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

$x{e}^x+C$

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

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

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

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

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

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