From Förberedande kurs i matematik 2

If we set \displaystyle u=2x+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 and \displaystyle du, which can give undesired effects. In this case, however, we have

\displaystyle du = (2x+3)'\,dx = 2\,dx

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

\displaystyle \begin{align} \int\limits_0^{1/2} e^{2x+3}\,dx &= \left\{\begin{align} u &= 2x+3\\[5pt] du &= 2\,dx \end{align}\right\} = \frac{1}{2}\int\limits_3^4 e^u\,du\\[5pt] &= \frac{1}{2}\Bigl[\ e^u\ \Bigr]_3^4 = \frac{1}{2}\bigl(e^4-e^3\bigr)\,\textrm{.} \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).