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=udu=x=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=2(u−1)eu+C.

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

exdx=2(x−1)ex+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.