Aus Online Mathematik Brückenkurs 2

The formula for partial integration reads

\displaystyle \int{f\left( x \right)}g\left( x \right)\,dx=F\left( x \right)g\left( x \right)-\int{F\left( x \right){g}'\left( x \right)\,dx},

where \displaystyle F\left( x \right) is a primitive function of \displaystyle f\left( x \right) and \displaystyle {g}'\left( x \right) is a derivative of \displaystyle g\left( x \right).

If we are to use partial integration, the integrand has to be divided up into two factors, a factor \displaystyle f\left( x \right) which we integrate and a factor \displaystyle g\left( x \right) which we differentiate. It is only when the product \displaystyle F\left( x \right){g}'\left( x \right) becomes simpler than \displaystyle f\left( x \right)g\left( x \right) that there is any point in partially integrating.

In the integral

\displaystyle \int{2xe^{-x}}\,dx,

it can seem appropriate to choose \displaystyle f\left( x \right)=e^{-x} and \displaystyle g\left( x \right)=2x, because then \displaystyle {g}'\left( x \right)=2 and we have only \displaystyle F\left( x \right)=-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}