Proving that $\int \frac{1}{x} \mathrm dx = \ln(|x|) + C_1$

Question

In all textbooks and online notes, there is always a table of antiderivatives and it always says $\int \frac {1}{x} \mathrm dx = \ln(|x|)+C_1$ but there is nowhere a proof. I found some proofs online but there is too much circular logic, assuming unproven hypothesis to reach the conclusion. Differentiating $\ln(x)$ can't be the most rigorous proof out there...

Answer 1

$$(\ln(|x|)+C)'=\ln'(|x|)=\frac{|x|'}{|x|}=\frac{\text{sgn}(x)}{|x|}=\begin{cases}\frac{1}{|x|}&if\ x>0\\\frac{-1}{|x|}&if\ x<0\end{cases}=\frac{1}{x}$$

Q.E.D.

Answer 2

Let $f(x)=\ln x$

Explicitly: $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac{\ln(x+h)-\ln(x)}h=\lim_{h\to0}\frac{\ln(1+h/x)}{h.(h/x)}.(h/x)=\frac1x$$

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EDIT: $$\frac{d}{dx}\ln|x|=\frac1{|x|}.\frac{d}{dx}|x|=\frac1x$$

Answer 3

Hint: For $h\neq0$, we have $~\displaystyle\int_1^xt^{~h-1}~dt=\bigg[\frac{t^h}h\bigg]_1^x=\frac{x^h-1}h.~$ For $h=0$, we have $~\displaystyle\int_1^x\frac{dt}t=$

$=\displaystyle\lim_{h\to0}\frac{x^h-1}h.~$ At the same time, $~\Big(a^x\Big)'=\displaystyle\lim_{h\to0}\dfrac{a^{x+h}-a^x}h=a^x~\lim_{h\to0}\dfrac{a^h-1}h.~$ Do you notice

anything “suspicious” ? ;-) Let e be the number for which this limit is $1$, and let $\ln=\log_{~\large e}.$ Then

it follows that $~\Big(a^x\Big)'=\Big(e^{x\ln a}\Big)'=(x\ln a)'\cdot e^{x\ln a}\cdot1=a^x\ln a.~$ Thus, $~\displaystyle\int_1^x\frac{dt}t=\ln x,$ which

for $x=e,~$ yields $~\displaystyle\int_1^{\large e}\frac{dt}t=1.~$ Now, let us inspect the numbers $u(h)$ for which $~\displaystyle\int_1^{u(h)}t^{h-1}~dt$

$=1.~$ Integrating, we have $\dfrac{u^h-1}h=1\iff u=\sqrt[^h]{1+h}.~$ Letting $h\to0,~$ we have $e=u(0)$

$=\displaystyle\lim_{h\to0}(1+h)^{^\tfrac1h}=\lim_{n\to\infty}\bigg(1+\dfrac1n\bigg)^n$. Hope this helps.

Answer 4

We will start by defining

$$\log(x) \equiv \int_1^x \frac{dt}{t}$$

and then showing that this is indeed the inverse function of the exponential function $e^{x}$. By definition and for $x>0$, $\log(x)$ is continious and monotonely increasing with $\log(1) = 0$.

We first start by showing that $\log(x) + \log(y) = \log(xy)$:

$$\log(x) + \log(y) \equiv \int_1^x \frac{dt}{t} + \int_1^y \frac{dt}{t} = \int_1^x\frac{dt}{t} + \int_x^{xy} \frac{d(t/x)}{(t/x)}= \int_1^{xy} \frac{dt}{t} \equiv \log(xy)$$

Where we switched the integration variables $t\to t/x$ in the last integral and used $\int_a^b + \int_b^c = \int_a^c$. By the same method we can show that $\log(x^n) = n\log(x)$:

$$\log(x^n) \equiv \int_1^{x^n} \frac{dt}{t} = \int_1^{x}\frac{ndt^{1/n}}{t^{1/n}} = n\int_1^x \frac{dt}{t} = n\log(x)$$

Since $\log(x)$ is monotonely increasing it has an inverse function $e(x)$ which satisfy

$$\log(e(x)) + \log(e(y)) = \log(e(x)e(y))\to e(x + y) = e(x)e(y)$$

Plugging in $x=y=0$ we have $e(0) = 1$. Further we find

$$\log(e(x)^{\frac{1}{x}}) = \frac{1}{x} \log(e(x)) \to e(x)^{\frac{1}{x}} = e(1)$$

so $e(x) = e(1)^x$. Finally, to show that $e(x) = e^x$ we must calculate $e(1)$ and show that this is really equal to $e$ ($ = 2.7128\ldots$). The definition of $e$ I'm going to use is that $e$ is the unique constant such that

$$\lim_{h\to 0} \frac{e^h-1}{h} = 1$$

since this is the definition that is used to show that $\frac{de^x}{dx} = e^x$. Plugging in $e(1)$ in the definition we get that our constant must satisfy

$$1 = \int_1^{e(1)}\frac{dt}{t}$$

and that $e(1) = e$ follows from Lucians answer above.

Answer 5

Cop-out answer

Define $\ln(x):=\int_1^x \frac{1}{t}dt$. Then the result follows immediately.

Serious answer

$$ \int \color{green}{\frac{1}{x}}dx = \int \color{green}{\sum_{k=0}^\infty (-1)^k (x-1)^k} dx \\$$ $$= \sum_{k=0}^\infty (-1)^k \int (x-1)^k dx \tag{pulling out the constant }\\ = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} (x-1)^{k+1} + C = \ln |x| +C $$

Answer 6

Here is the proof I wrote with Ian's guiding. Ian, thank you for pointing me in the right direction for what I was looking for; $ln(x)$ is defined to be the inverse function of $e^x$. The inverse function theorem states that $(f^{-1})'(b)=\frac{1}{(f'(a))}$, $b=f(a)$. Let $f(x)=\ln(x), f^{-1}(f^{-1}(x))=f(x)$, thus, $f^{-1}(x)=e^x$. $(f^{-1})'=(e^x)'=e^x$. Ergo, on the LHS, $(f^{-1})'(f(x))=e^{\ln(x)}=x$. RHS then becomes $\frac{1}{(\ln(x))'}$, thus $x=\frac{1}{(\ln(x))'} \iff x^{-1}=(\frac{1}{(\ln(x))'})^{-1} \iff \frac{1}{x}=(\ln(x))' \iff \int \frac{1}{x} dx= \int( \ln(x))' dx$. Using the Fundamental theorem of calculus, $\int \frac{1}{x} dx=\ln(x)+c$.Then suspect c is $iπ$, and because of the identity $\ln(x)=\ln(-x)+iπ$, $\ln(x)$ becomes $\ln(|x|)$, but I don't know how to prove that.