I know that $\int \frac{1}{x}~dx = \ln|x| + C$ and I know the antiderivative method works for all powers of $x$ except $-1$. But why is that the case? I am still in high school and teachers aren't really helpful with these questions.

Edit: I do realize dividing by zero is meaningless. That's not the question. Just because dividing by zero is meaningless doesn't mean mathematicians just choose to ignore it and work just make up a new answer. There has to be an explanation, and that's what I am asking for. The explanation.

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    $\begingroup$ What would $\frac{x^0}{0}$ mean? $\endgroup$
    – egreg
    Jan 28 '17 at 17:29
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    $\begingroup$ @egreg Well, that's undefined. But just because something's undefined doesn't mean they can just avoid it and choose not to work with it. I just want to mathematically understand why doesn't it work. $\endgroup$
    – Eyad H.
    Jan 28 '17 at 17:32
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    $\begingroup$ People don't avoid $\dfrac{x^0}0$ as the antiderivative of $1/x$ because $\dfrac{x^0}0$ is undefined. It's avoided because it doesn't satisfy the definition of antiderivative when we apply the definition to $1/x$. $\endgroup$
    – user307169
    Jan 28 '17 at 17:38
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    $\begingroup$ @Eyad H. Just because what you write is undefined, the mathematicians choose not to work with it and that's exactly what undefined things means. $\endgroup$ Jan 28 '17 at 17:38
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    $\begingroup$ Since $x^0/0$ isn't defined it can't be an antiderivative of anything. But if we decide to DEFINE as NOTATION $x^0/0 := \lim_{h\rightarrow} \frac {x^h}h$ then indeed $\int \frac 1x dx $ *does* indeed equal $x^0/0 = \lim \frac{x^h}h = \ln x + 1$. $\endgroup$
    – fleablood
    Feb 23 '17 at 17:18

I think it is important to understand how the formula $$\int x^{n}\,dx = \frac{x^{n + 1}}{n + 1} + C\tag{1}$$ is coming into picture. Well, it is a result of inverting the differentiation formula $$\frac{d}{dx}x^{n + 1} = (n + 1)x^{n}\tag{2}$$ If $n + 1\neq 0$ then we can divide the formula $(2)$ by $(n + 1)$ to get $$\frac{d}{dx}\frac{x^{n + 1}}{n + 1} = x^{n}\tag{3}$$ If we put $n = -1$ then the formula $(2)$ remains valid (check!) but the division by $(n + 1)$ is not allowed and hence the formula $(3)$ becomes meaningless and therefore it can not be inverted to get the integral formula $(1)$.

So it is not possible to invert differentiation formula like $(2), (3)$ to obtain $(1)$ in case $n = -1$. The natural question to ask now is:

What is the anti-derivative of $x^{-1}$ and how to figure it out if the differentiation formulas like $(2), (3)$ don't offer any help?

However before answering the above question, it makes sense to ask one further question:

What grounds do we have for thinking about the anti-derivative of $x^{-1}$? Perhaps there might not be an anti-derivative for this particular function.

This is where the Fundamental Theorem of Calculus (FTC) comes into picture which provides us with the following:

Corollary of FTC: Let $f$ be continuous on $[a, b]$. Then there is an anti-derivative $F$ of $f$ such that $F'(x) = f(x)$ for all $x\in [a, b]$. Moreover any two anti-derivatives of $f$ differ by a constant.

And the above is true even if $[a, b]$ is replaced by $(a, b)$.

The function $f(x) = x^{-1}$ under consideration is continuous on intervals $(-\infty, 0)$ and $(0, \infty)$ and hence on each of these intervals it does possess an anti-derivative.

The FTC also gives us an expression for the anti-derivative in terms of what is usually called a definite integral (or more technically a Riemann integral). Thus if $x > 0$ then the following integral $$F(x) = \int_{1}^{x}\frac{dt}{t}$$ is an anti-derivative of $f(x) = 1/x$ in interval $(0, \infty)$.

In case the presentation of calculus is such that anti-derivatives are studied before Riemann integral (this is the more common approach used in introductory calculus courses), then it is sufficient to assume (via the use of corollary of FTC mentioned above) that there exists an anti-derivative $F(x)$ of $f(x) = 1/x$ for $x > 0$ with the following properties $$F(1) = 0, F(xy) = F(x) + F(y), F(x/y) = F(x) - F(y), F'(x) = 1/x\tag{4}$$ for all positive $x, y$. The function $F$ introduced above with some nice / curious properties is traditionally denoted by $\log x$ (or by $\ln x$) and is famously known as the logarithm function. It is now easy to prove that $\log (-x)$ is anti-derivative of $1/x$ for $x < 0$ and thus we have the formula $$\int \frac{dx}{x} = \log |x| + C\tag{5}$$

Looking at the formulas $(1)$ and $(5)$ an inquiring mind may ask if there is any connection between these formulas. There is a connection between the formulas $(1)$ and $(5)$ but it is visible only when we switch to definite integrals and the connection has been amply described in the answer from user "Simply Beautiful Art". But unless one has studied definite integrals it is best to think of $(1)$ and $(5)$ as separate formulas.


Notice that


but at $n=-1$, we get division by $0$, and clearly this is bad. However, following this line of logic, we can take the limit as $n\to-1$ to get


You might recognize this better if I remind you of the following limit:


And combine all this, you'll get


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    $\begingroup$ I did not get how $\lim_{h\to0}\frac{x^h-1}h$ and $ 1=\lim_{h\to0}\frac{e^h-1}h$ will give the last step. Forgive my ignorance. $\endgroup$
    – user312097
    Jan 28 '17 at 18:51
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    $\begingroup$ @A---B It's no problem. We have the following, with $h'=h\ln(x)$.$$\lim_{h\to0}\frac{x^h-1}h=\ln(x)\lim_{h\to0}\frac{e^{h\ln(x)}-1}{h\ln(x)}=\ln(x)\lim_{h'\to0}\frac{e^h-1}h$$ $\endgroup$ Jan 28 '17 at 18:53
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    $\begingroup$ The easy answer is what others gave: to find an antiderivative of a function $f$, find a function whose derivative is $f.$ But that explanation leaves a puzzling gap in the power law, which you've just filled. And this gap is something I've just been putting up with for nearly four decades! $\endgroup$
    – David K
    Jan 28 '17 at 19:37
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    $\begingroup$ The usual way works fine, which makes it possible to tolerate it so long. It just isn't so neat and symmetrical. $\endgroup$
    – David K
    Jan 28 '17 at 19:48
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    $\begingroup$ @A---B Alternatively, L'Hôpital. $\endgroup$ Feb 8 '17 at 19:23

Completely setting aside the meaninglessness of $\dfrac{x^0}0$ (recall that dividing by zero is a no-no)..

It's because the derivative of $\dfrac{x^0}0$ is not $\dfrac1x$. Its derivative isn't even defined since the expression $\dfrac{x^0}0$ itself isn't defined.

The derivative of $\ln x$ is $\dfrac1x$.

Therefore the antiderivative (by definition of antiderivative) of $\dfrac1x$ is $\ln x$ (plus an arbitrary constant).


Notice that, because integration involves addition of a constant, the following is also an antiderivative of $x^n$: $$\frac{x^{n+1}-1}{n+1}$$ It is indeterminate at $n=-1$, but one might instead consider $$\lim_{n\to -1}\frac{x^{n+1}-1}{n+1}$$ which, by L'Hopital's rule, is equal to $$\begin{align} \lim_{n\to -1}\frac{x^{n+1}-1}{n+1} &=\lim_{n\to -1}\frac{e^{(n+1)\ln(x)}-1}{n+1}\\ &=\lim_{n\to -1}\frac{e^{(n+1)\ln(x)}\ln(x)}{1}\\ &=e^0\ln(x)\\ &=\ln(x)\\ \end{align}$$

  • $\begingroup$ This is wrong, you end up with 1/0 for most values of $x$, and L'Hopital's rule is invalid in such a case. $\endgroup$ Oct 6 '18 at 18:00
  • $\begingroup$ @SimplyBeautifulArt I fixed it. Better? My apologies, I wrote this answer a long time ago. :P $\endgroup$ Oct 6 '18 at 18:03
  • $\begingroup$ :P Also check out my chatroom sometime! (I've got an interesting problem you might enjoy) $\endgroup$ Oct 6 '18 at 18:17

Since you are in high school, I just give you some "unrigorous" argument. First, you can show that $x\longmapsto \frac{1}{x}$ is integrable on all $[a,b]$ that doesn't contain $0$ (since all continuous function on $[a,b]$ is integrable on $[a,b]$). Let consider $I=]0,\infty [$. Let $x>0$ and $0<a<x$. You know that $$\left(\int_a^x\frac{1}{t}\mathrm d t\right)'=\frac{1}{x}.$$ But you also know that $(\ln(x))'=\frac{1}{x}$. A famous result says that if $f'=g'$, then $f=g+C$ for a constant $C$. We conclude that $$\int_a^x\frac{1}{t}\mathrm d t=\ln(x)+C.$$


In a sense it is.

$\lim_{h\rightarrow 0} \frac {x^h}{h}= \ln x + 1$

So $\int \frac 1x dx = \ln x + 1 + C = \ln x + C$.

$\int x^n = \frac {x^{n+1}}n +C$ isn't magic. It is the answer to "what function $F$ is such that $F'(x) =\lim \frac {F(x+h) - F(x)}h = f(x)$?"

Since $\frac {x^0}0$ isn't any function at all there can't be any function with it as an antiderivative.

However for all $k \ne 0$ $\frac {x^k}k$ is a function and it does follow such that $\int x^{k-1} = \frac {x^k}{k} + C$.

Which raises the question: Well gosh, As long as we are not evaluating at $x = 0$ then function $\frac 1x$ is perfectly well behaved so it ought to have an antiderivative; what is it?

And we simply define, out of whole cloth that it is, by definition a function we call $\ln x = \int_{1}^x \frac 1x dx$.

And although we don't usually do this, we can show that $\lim_{h\rightarrow 0} \frac {x^h}{h}= \ln x + 1$ so that could have been our definition instead.

(What we usually do instead is prove $\lim_{n\rightarrow \infty} (1 + \frac 1n)^n$ exists as a real number. We define that number as $e$ and show that $\ln x = \log_e x$).

(Actually what we do is a bit more roundabout as we have to define what $\log_b$ means and we have to define what $b^x; x \in \mathbb R$ means first. Which usually involves finding the value of $e$ as a limit or the function $\ln x$ as a limit.)

  • $\begingroup$ "$\lim_{h\rightarrow 0} \frac {x^h}{h}= \ln x + 1$" ? $\endgroup$
    – Winther
    Oct 6 '18 at 18:10
  • $\begingroup$ Apparently I meant $\lim_{h\to 0} \frac {x^h - 1}h = \ln x$. Not sure why I consistantly wrote it as the above. $\endgroup$
    – fleablood
    Oct 6 '18 at 19:23

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