Hans Georg Schaathun
Oktober 2016
$$\int (2x+1)e^{2x}dx = $$
$$\frac{d}{dx}\big[f(x)\cdot g(x)\big] = f'(x)g(x) + f(x)g'(x)$$
$$f(x)\cdot g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx$$
$$\int f(x)g'(x)dx = f(x)\cdot g(x) - \int f'(x)g(x)dx$$
$$\int (2x+1)e^{2x}dx = \int f(x)g'(x)dx$$
$$f(x) = 2x+1 \quad g(x) = \frac12e^{2x}$$
$$\int (2x+1)e^{2x}dx = (2x+1)\frac{e^{2x}}{2} - \int 2\cdot\frac{e^{2x}}2dx$$
$$\int (2x+1)e^{2x}dx = \big(x+\frac12\big)e^{2x} - \frac12\cdot e^{2x}=xe^{2x}$$
$$\int f'(x)g(x)dx = f(x)\cdot g(x) - \int f(x)g'(x)dx$$
$$\int U'Vdx = UV - \int UV'dx$$
\( U' = \frac{dU}{dx} \)\(\quad\Longrightarrow\quad U'dx = dU \)
\( V' = \frac{dV}{dx}\quad\Longrightarrow\quad V'dx = dV \)
$$\int VdU = UV - \int UdV$$