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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....

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    $\begingroup$ You said $du=2x$ but actually it should be $du=2xdx$ and $xdx=\frac{1}{2}$ also doesn't make sense. $\endgroup$ – user137035 Sep 11 '14 at 7:14
  • $\begingroup$ 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. $\endgroup$ – rubik Sep 11 '14 at 7:27
  • $\begingroup$ @rubik Edits are not intended for this type of change. $\endgroup$ – AlexR Sep 11 '14 at 7:33
  • $\begingroup$ It should be $du = 2xdx, xdx = \frac{1}{2}du$ sorry! $\endgroup$ – rablentain Sep 11 '14 at 7:36
  • $\begingroup$ @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. $\endgroup$ – rubik Sep 11 '14 at 7:53
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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: $$\frac{1}{2}\ln{|u|}\ =\, \frac{1}{2}\ln{(x^2+1)}.$$

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    $\begingroup$ For the sake of correctness, $\int \frac1u \mathrm du = [\ln |u|]$. (Note the absolute value). The final result is correct, though. $\endgroup$ – AlexR Sep 11 '14 at 7:27
  • $\begingroup$ @AlexR: You're right. Thank you. $\endgroup$ – SPK.z Sep 11 '14 at 7:31
  • $\begingroup$ I get it now, thanks! if I write it like $int\frac{xdx}{x²+1}$ I found it easier to understand... $\endgroup$ – rablentain Sep 11 '14 at 7:43
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With $u=x^2+1$ and thus $du=2x dx$ you get $$\int\frac{x}{x^2+1}dx=\frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}\ln|u| = \frac{1}{2}\ln(1+x^2)$$

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  • $\begingroup$ For the sake of correctness, $\int \frac1u \mathrm du = [\ln |u|]$. (Note the absolute value). The final result is correct, though. $\endgroup$ – AlexR Sep 11 '14 at 7:28

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