# Thread: Du Smaller than Dx Integration Problem

1. ## Du Smaller than Dx Integration Problem

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.

2. ## Re: Du Smaller than Dx Integration Problem

You started to make a u-substitution, but you should have $\displaystyle du=dx$ instead of $\displaystyle du=1$. If you solve for x in terms of u, you get x=u-1. So you have everything you need to make the substitution:

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

It's a little easier to see how to proceed now.

The book doesn't make the substitution. It just uses x+1 "as is" instead of u. But the steps are the same.

Hope this helps,
Hollywood