Q:

Confusing step in WolframAlpha integration

The problem was: ∫2e^(-x/4)dx
And the steps WolframAlpha used were:
1.) Take the integral ∫2e^(-x/4)dx
2.) Factor out constants 2∫e^(-x/4)dx
3.) For the integrand e^(-x/4), substitute u = (-x/4) and du = -1/4 dx : -8∫e^(u)du
The rest I understand, but where does the -1/4 dx come from?

A:

From the Fundamental Theorem of Calculus.
$F(x) = \int_a^{u(x)} f(t) dt\\
F'(x) = f(u(x)) u'(x)$
Operating in reverse.
If you want to integrate some function that has the right "shape."
$\int f(u(x)) u'(x) dx$
i.e $\int e^{-\frac {x}{4}}(-\frac 14 dx)$
then you can say $u(x) = -\frac {x}{4}, u'(x) = -\frac {1}{4}$
although we usually write it as $du = -\frac 14 dx$
$\int e^{u} du = e^u + C$
and then reverse the substitution.
$e^{-\frac x4} + C$
But sometimes you get something that is not so perfect.
$\int e^{-\frac {x}{4}} dx$
then you do a little trickery to get the right form
$\int e^{-\frac {x}{4}}(-4)(-\frac 14) dx\\
-4\int e^{-\frac {x}{4}}(-\frac 14 dx)\\$
and substitute.

