Integral of $\int 1/x dx$

Question

Is this a sufficient proof for this integral?:

$$\int \frac{dx}{x} = \ln |x| + \mathcal{C}$$

Let $$x = e^{u} : $$

$$\int \frac{dx}{x} = \int du = u + \mathcal{C} = \ln |x| + \mathcal{C}$$

I'm not sure :S I don't know if my logic's a bit wishy washy/ circular with this :S And when I've looked, most websites seem to say it's a definition rather than a result but...I guess I just want a proof for this. Anyone have any ideas/ can validate whether the above is correct?

Thanks very much

I take it you are trying to prove $\int(dx/x)=\log x+C$. But this follows from the derivative of $\log x$ being $1/x$ --- are you not allowed to use that? — Gerry Myerson, Jun 11 '13 at 13:06
Yes, sorry, that's what I was trying to prove yes. DOH! So if I can prove the derivative of ln x is 1/x then it follows that the anti-derivative of 1/x is lnx + c? I think I remember something like that...the uniqueness theorem of anti-derivatives or something? — user78416, Jun 11 '13 at 13:07
your proof above seems logical enough though...i mean at least i don't see why it wouldn't be..hmm — Sidd Singal, Jun 11 '13 at 13:09
Your proof is right, there is a small mistake though. The substitution $x=e^u$ only works for positive $x$. You need to treat the case $x$ negative separately. — N. S., Jun 11 '13 at 13:15
The Fundamental Theorem Of Calculus (alternatively, the definition of $\int f(x),dx$). — Gerry Myerson, Jun 12 '13 at 09:03

score 5 · Answer 1 · answered Jun 11 '13 at 13:17

5

Your proof is valid but you can add more precision:

if $x\in(0,+\infty)$ we pose $x=e^u$ and we find $$\int \frac{dx}{x} = \ln x+ \mathcal{C}$$
if $x\in(-\infty,0)$ we pose $x=-e^u$ and we find $$\int \frac{dx}{x} = \ln (-x) + \mathcal{C}$$ then we conclude.

answered Jun 11 '13 at 13:17

Ah, excellent. Thanks very much! – user78416 Jun 11 '13 at 13:51
You're welcome. – Jun 11 '13 at 13:52
why $x = e^{u}$ and $x = -e^{u}$ what's the purpose? :) – EM4 Jan 21 '21 at 16:08

score 3 · Answer 2 · answered Jun 11 '13 at 13:11

3

Looks fine to me. By Taylor's Theorem, you can also write

$$\frac{1}{x} = \sum_k (-1)^k (x-1)^k$$

and integrate in the radius of convergence term by term, recognizing the result on the right-hand side. Assuming $x > 0$,

$$\begin{split} \int \frac{dx}{x} &= \int \sum_{k=0}^\infty (-1)^k (x-1)^k dx \\ &= \sum_{k=0}^\infty (-1)^k \int (x-1)^k dx \\ &= \sum_{k=0}^\infty \frac{(-1)^k}{k+1} (x-1)^{k+1} + C \\ &= \ln x +C \end{split} $$

answered Jun 11 '13 at 13:11

gt6989b

54,422

That's an interesting approach! Thanks – user78416 Jun 11 '13 at 13:51
1

@user78416 the problem is that to "recognize" the series for the $\ln x$ in the last step, you need to know that $(\ln x)' = 1/x$ which is the crux of your question :). But it does approach it from a slightly different side... – gt6989b Jun 11 '13 at 13:54
@gt6989b, the absolute value is missing in $\ln(x)$ above. – Amad27 Oct 09 '14 at 13:29
@Amad27 not really, since we are assuming $x>0$... – gt6989b Nov 10 '19 at 17:15

Lisa · Answer 3 · 2013-06-11T13:44:30.513

0

You solved the equation correctly (with the small negligence for negative x, which has been corrected by Sami).

If you want to (or are allowed to) use

$$ \frac{\partial}{\partial x} ln(x) = \frac{1}{x}, $$

then you might be interested in this proof.

edited Jun 11 '13 at 13:44

answered Jun 11 '13 at 13:39

Lisa

853

Integral of $\int 1/x dx$

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