Lösung 2.3:2d

Aus Online Mathematik Brückenkurs 2

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=xlnx−xx1dx=xlnx−1dx=xlnx−x+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=ueu−1eudu=ueu−eudu=ueu−eu+C=u−1eu+C

and the answer becomes

lnxdx=lnx−1elnx+C=lnx−1x+C