Lösung 2.3:2a

Aus Online Mathematik Brückenkurs 2

Had the integral instead been

ex12xdx 

it is quite obvious that we would substitute u=x , but we are missing a factor 12x which would take account of the derivative of u which is needed when dx is replaced by du. In spite of this, we can try the substitution u=x  if we multiply top and bottom by what is missing:

exdx=ex2x12xdx=u=xdu=xdx=12xdx=eu2udu

Now, we obtain instead another, not entirely simple, integral, but we can calculate the new integral by partial integration ( 2u is the factor that we differentiate and eu is the factor that we integrate),

eu2udu=eu2u−eu2du=2ueu−2eudu=2ueu−2eu+C=2u−1eu+C

If we substitute back u=x , we obtain the answer

exdx=2x−1ex+C 

As can be seen, it is possible to mix different integration techniques and often we need to experiment with different approaches before we find the right one.