Du Smaller than Dx Problem

$\displaystyle \int \dfrac{2x}{x + 1}dx$

$\displaystyle u = (x + 1)$

$\displaystyle du = 1$

Reasoning Behind the Problem (according to the book):

$\displaystyle \int \dfrac{2x}{x + 1}dx = 2 \int \dfrac{x}{x + 1}dx$

$\displaystyle = 2 \int (\dfrac{x + 1}{x + 1} - \dfrac{1}{x + 1})dx$

$\displaystyle = 2 \int (1 - \dfrac{1}{x + 1})dx$

$\displaystyle = 2 \int dx - 2 \int \dfrac{dx}{x + 1}$

$\displaystyle \int \dfrac{2x}{x + 1}dx = 2x - 2 \ln(x + 1) + C$ Answer

But have no clue what's going on (in any of it). Any hints? Of course I understand what to do if du = dx at the start. In that case, you do nothing. However, if du is bigger than dx than you multiply the dx by a constant, and then balance it with the reciprocal (of the constant) in the answer.