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

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....

• You said $du=2x$ but actually it should be $du=2xdx$ and $xdx=\frac{1}{2}$ also doesn't make sense.
– user137035
Commented Sep 11, 2014 at 7:14
• I quote @i.ozturk, from $\mathrm du = 2x\mathrm dx$ you get $x\mathrm dx = \frac{\mathrm du}{2}$. I proposed an edit with the corrections. Commented Sep 11, 2014 at 7:27
• @rubik Edits are not intended for this type of change. Commented Sep 11, 2014 at 7:33
• It should be $du = 2xdx, xdx = \frac{1}{2}du$ sorry! Commented Sep 11, 2014 at 7:36
• @AlexR: Oh ok. I was suspecting that I couldn't change the post like that, but I tried anyway. Thank you for clarifying it to me. Commented Sep 11, 2014 at 7:53

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: $$\frac{1}{2}\ln{|u|}\ =\, \frac{1}{2}\ln{(x^2+1)}.$$
• For the sake of correctness, $\int \frac1u \mathrm du = [\ln |u|]$. (Note the absolute value). The final result is correct, though. Commented Sep 11, 2014 at 7:27
• I get it now, thanks! if I write it like $int\frac{xdx}{x²+1}$ I found it easier to understand... Commented Sep 11, 2014 at 7:43
With $u=x^2+1$ and thus $du=2x dx$ you get $$\int\frac{x}{x^2+1}dx=\frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}\ln|u| = \frac{1}{2}\ln(1+x^2)$$
• For the sake of correctness, $\int \frac1u \mathrm du = [\ln |u|]$. (Note the absolute value). The final result is correct, though. Commented Sep 11, 2014 at 7:28