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!
 A: $du = 2 \, dx$ 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}
A: For the integral $3\int_e^{2x}dx$ we can subtitute $u=2x$. Then the derivative of u with respect to x is $2$, so the differntial $du=2dx$ or $\frac{1}{2}du=dx$. Now we can substitute these values into the integral to get:
$$3\int e^{2x}dx=3\int e^u \frac{1}{2}du=\frac{3}{2}\int e^u du$$
All we did was make the substitution and factor out the constant. 
A: Please forgive me as this in essence the same answer as the others but I wanted to illustrate something subtle.
You have $$3\int e^{2x}\;dx$$
Now before I substitute anything I can rearrange the integral.
$$3\int \frac{1}{2}e^{2x} \times2\;dx $$
Now I substitute $u=2x\implies \frac{du}{dx}=2$
This leaves the integral as $$\frac{3}{2}\int e^{u}\times\frac{du}{dx}\;dx$$
Notice the constants can move through the integral sign.
Now we note the equivalent operators $\int ...\frac{du}{dx}\;dx\equiv\int...dx$  
I think of this as the reverse of implicit differentiation and it is a consequence of the chain rule.
So we have $$\frac{3}{2}\int e^{u}\;du$$
I wanted to highlight that the substitution worked because we could rearrange the original integrand in just the right way.  
The $\frac{du}{dx}$ term acts as a scale factor and in multivariable calculus is called the Jacobian.
