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I saw someone write this for showing substitution. Is it correct.

$$\int \frac{2x}{\sqrt{x^2+2}}\, \mathrm{d}x$$

$$\int \frac{\mathrm{d}u}{\sqrt{u}}\, \mathrm{d}\left(2xdx\right)$$

Just wondering about the various methods to write an integral, because I have seen a couple of different methods, but nothing like this before.

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That second form looks extremely dodgy. –  J. M. Apr 9 '13 at 14:03
    
@J.M. That is what I was thinking, but I am not an expert. So is it valid? –  yiyi Apr 9 '13 at 14:05
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Remove the meaningless $d(2xdx)$ and then the formula will be correct, given $u=x^2+2$. –  1015 Apr 9 '13 at 14:14
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@MaoYiyi The second notation is a mess - it doesn't mean anything. There is no such think as $d(2xdx)$ in any format, and there should only be one $dZ$ in a simple integral. So no, this is not a valid - the second formula is not only wrong, it is meaningless. –  Thomas Andrews Apr 9 '13 at 14:16
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Because of a medical problem (a mild allergy to fractions) I would prefer to let $u^2=x^2+2$. Then $2x\,dx=2u\,du$ and when the smoke clears we are finding $\int 2\,du$. –  André Nicolas Apr 9 '13 at 16:04
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The second integral is nonsensical - it does not mean anything. Thus it is worse than wrong. The most egregious problem is $d(2xdx)$. That simply means nothing, and if you delete all of those characters, the substitution is correct.

Indeed, I suspect the writer wanted to say that when $u=x^2+2$, $du=2xdx$, but got tangled up in the formatting.

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What is this "valid" you keep mentioning? The first and last steps are $$ \int \frac{2x}{\sqrt{x^2+2}}\, \mathrm{d}x $$ and $$ \int\frac{du}{\sqrt{u}} $$ If there is a calculation $u=x^2+2, du=2xdx$ in there somewhere, OK. But it looks like the "someone" wrote his notes all mixed together. If he gets to the last step anyway, great. But if not, maybe recommend that he not combine everything together like that.

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I thought my classmate was wrong, but I am not great at calculus so i asked here. –  yiyi Apr 9 '13 at 14:31
    
In a published paper, of course you wouldn't write that. In your scribbles you make as you work on a problem, maybe you would write that. If it gets you to the proper answer, I have no objection to it. –  GEdgar Apr 9 '13 at 14:33
    
GEdgar, did you see the $d(2xdx)$? That's nonsensical in any format. Perhaps the fellow student just copied the text sloppily off the blackboard. –  Thomas Andrews Apr 9 '13 at 14:54
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No need to substitute: if $\,f\,$ is a diffentiable function then

$$\int\frac{f'}{\sqrt f} dx=2\sqrt f+ K$$

In this case , $\,2x=(x^2+2)'\,$ , so directly

$$\int\frac{2x}{\sqrt{x^2+2}}dx=\int\frac{d(x^2+2)}{\sqrt{x^2+2}}=2\sqrt{x^2+2}+K$$

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I know that rule, however, to show the steps of why that is correct one has to do what I have seen named as "u-substitution". Is the method listed valid? –  yiyi Apr 9 '13 at 14:07
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To be fair, that rule just is an instance of substitution. –  Thomas Andrews Apr 9 '13 at 14:12
    
@ThomasAndrews is what is written in the question valid method for showing substitution? –  yiyi Apr 9 '13 at 14:14
    
No @MaoYiyi: what is written is a valid method to avoid substitution and weird things like the one you wrote you saw in yout post, which is in the very frontier between weird looking and wrong. The method I show has, imvfho, the advantage that you don't have to justify the substitution is by a function that is continuously differentiable, injective, positive or whatever other condition one might be required to produce. –  DonAntonio Apr 9 '13 at 14:49
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@DonAntonio You do realize the question wasn't about how to solve the problem, but that OP was confused by the above notation? "Just wondering about the various methods to write an integral, because I have seen a couple of different methods, but nothing like this before." –  Thomas Andrews Apr 9 '13 at 14:58
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