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
lnx
We integrate the factor
1
lnxdx=x
lnx−
x
x1dx=x
lnx−
1dx=x
lnx−x+C
Method 2 (substitution)
It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression
lnx
dx=x1dx
and because
dx=eudu
Thus, the substitution becomes
lnxdx=
u=ln xdx=eudu
=
ueudu
Now, we carry out a partial integration
ueudu=u
eu−
1
eudu=u
eu−
eudu=u
eu−eu+C=
u−1
eu+C
and the answer becomes
lnxdx=
lnx−1
elnx+C=
lnx−1
x+C