Aus Online Mathematik Brückenkurs 2

If we set \displaystyle u=\text{2}x+\text{3} , the integral simplifies to \displaystyle e^{u}. However, this is only part of the truth. We must in addition take account of the relation between the integration element \displaystyle dx\text{ } and \displaystyle du, which can give undesired effects. In this case, however, we have

\displaystyle du=\left( \text{2}x+\text{3} \right)^{\prime }\,dx=2\,dx

which only affects by a constant factor, so the substitution \displaystyle u=\text{2}x+\text{3} seems to work, in spite of everything:

\displaystyle \begin{align} & \int\limits_{0}^{{1}/{2}\;}{e^{\text{2}x+\text{3}}}\,dx=\left\{ \begin{matrix} u=\text{2}x+\text{3} \\ du=2\,dx \\ \end{matrix} \right\}=\frac{1}{2}\int\limits_{3}^{4}{e^{u}\,du} \\ & =\frac{1}{2}\left[ e^{u} \right]_{3}^{4}=\frac{1}{2}\left( e^{4}-e^{3} \right) \\ \end{align}

NOTE: Another possible substitution is \displaystyle u=e^{2x+3} which also happens to work (usually, such an extensive substitution almost always fails).