How do I handle $dx$ in $u$-substitution?

Question

I am having trouble with an example of $u$-substitution:

$$ \int \frac{x}{x²+1}dx$$

In the next step they write:

Let $u = x^2+1$ which seems like a good choice. Then $du = 2x$ and that is also obvious, but then: $xdx = \frac{1}{2}$ and then use that to do something as I don't understand.

How is the $xdx$ value calculated and what are they using it for? As you see, I am having trouble with understanding $u$-substitution in general....

Answer 1

If you choose $u=x^2+1$, then taking the derivative with respect to $x$ gives: $$\frac{\textrm{d}u}{\textrm{d}x}=2x.$$ Therefore $\textrm{d}u=2x\textrm{d}x$ or $x\textrm{d}x=\textrm{d}u/2$. Now your integral is written: $$\frac{1}{2}\int\frac{1}{u}\textrm{d}u,$$ which is easier to solve than the original integral. It is equal to: $$\frac{1}{2}\ln{|u|}\ =\, \frac{1}{2}\ln{(x^2+1)}.$$

Answer 2

With $u=x^2+1$ and thus $du=2x dx$ you get $$\int\frac{x}{x^2+1}dx=\frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}\ln|u| = \frac{1}{2}\ln(1+x^2)$$