How we get $\frac{3}{2} \int e^u du$ from $3 \int e^{2x} dx$ I cannot understand how we get $\frac{3}{2} \int e^u du$ from   $3 \int e^{2x} dx$ after substituting $u = 2x$ and $du = 2dx$. Wolfram Pro gives this substitution step never explaining how we get $\frac{3}{2}$. Thanks!

$du = 2 \, dx$ means $dx = \dfrac{1}{2}du$.  So then
\begin{align}
  3\int \underbrace{e^{2x}}_{e^u} \ \ \underbrace{dx}_{\frac{1}{2} \, du}
    &= 3 \int e^u \frac{1}{2} \, du\\[0.3cm]
    &= \frac{3}{2} \int e^u \, du
\end{align}
