Comment by ajkjk
7 hours ago
well.. no, not exactly. If u = u(x) then du = u'(x) dx holds rigorously, and then you can substitute du/u' = dx in an integral.
7 hours ago
well.. no, not exactly. If u = u(x) then du = u'(x) dx holds rigorously, and then you can substitute du/u' = dx in an integral.
I'm thinking more along the lines of knocking a '2x' out of an integral from d/dx of like 2x^2.