Lösung 2.3:1a

Aus Online Mathematik Brückenkurs 2

The formula for partial integration reads

fxgxdx=Fxgx−Fxgxdx ,

where Fx  is a primitive function of fx  and gx  is a derivative of gx .

If we are to use partial integration, the integrand has to be divided up into two factors, a factor fx  which we integrate and a factor gx  which we differentiate. It is only when the product Fxgx  becomes simpler than fxgx  that there is any point in partially integrating.

In the integral

2xe−xdx ,

it can seem appropriate to choose fx=e−x  and gx=2x , because then gx=2  and we have only Fx=−e−x  left,

\displaystyle \begin{align} & \int{2xe^{-x}}\,dx=2x\centerdot \left( -e^{-x} \right)-\int{2\centerdot }\left( -e^{-x} \right)\,dx \\ & =-2xe^{-x}+2\int{e^{-x}\,dx} \\ \end{align}

It remains only to integrate \displaystyle e^{-x} and we are finished:

\displaystyle \begin{align} & =-2xe^{-x}+2\left( -e^{-x} \right)+C \\ & =-2xe^{-x}-2e^{-x}+C \\ & =-2\left( x+1 \right)e^{-x}+C \\ \end{align}