Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.

No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message


jsMath

Lösung 2.3:2d

Aus Online Mathematik Brückenkurs 2

Wechseln zu: Navigation, Suche

We shall solve the exercise in two different ways.

Method 1 (partial integration)

As first sight, partial integration seems impossible, but the trick is to see the integrand as the product


1lnx


We integrate the factor 1 and differentiate lnx,


1lnxdx=xlnxxx1dx=xlnx1dx=xlnxx+C


Method 2 (substitution)

It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression u=ln x. The problem we encounter is how we should handle the change from dx to du. With this substitution, the relation between dx and du becomes


du=lnxdx=x1dx 


and because u=ln x, then x=eu and we have that


du=1eudxdx=eudu


Thus, the substitution becomes


lnxdx=u=ln xdx=eudu=ueudu 


Now, we carry out a partial integration


ueudu=ueu1eudu=ueueudu=ueueu+C=u1eu+C


and the answer becomes


lnxdx=lnx1elnx+C=lnx1x+C