Integrals of $1/x$ and $1/x^2$

Question

Why is the integral of $1/x$ is $\ln|x|$, while the integral of $1/x^2$ is not $\ln|x^2|/(2x)$?

I figured that it has something to do with the power of $-1$ because if you add $+1$ to $-1$ it will be $0$ and you can not divide to zero. But this is just a theory.

Your theory is, loosely speaking, correct. All real powers can be antidifferentiated using the power law except $-1$. For $-1$, the power law fails because the derivative of $x^0$ is just $0$. We often define $\ln$ to be the antiderivative of $-1$, and such that $\ln(1)=0$. — FShrike, Jan 23 '22 at 10:49
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Jan 23 '22 at 10:49
There was a similar question yesterday: https://math.stackexchange.com/questions/4363175/why-does-int-frac1-sqrt-udu-neq-ln-sqrt-u-hold-true#comment9112623_4363175 — Gary, Jan 23 '22 at 10:51
It may help you https://math.stackexchange.com/questions/417417/integral-of-int-1-x-dx — Lion Heart, Jan 23 '22 at 10:53

miracle173 · Answer 1 · 2022-01-23T11:11:56.910

2

Why should it? $$\left(\ln|x|\right)'=\frac 1 x$$ $$\left(\frac{\ln|x^2|}{2x}\right)'=\frac{1}{{{x}^{2}}}-\frac{\ln{(x)}}{{{x}^{2}}}\ne\frac 1 {x^2}$$

edited Jan 23 '22 at 11:11

answered Jan 23 '22 at 11:04

miracle173

11,049

score 2 · Answer 2 · answered Jan 23 '22 at 11:10

It looks like you are assuming the following rule: if $F$ is an antiderivative of $f$, and $g'$ is the derivative of $g$, then an antiderivative of $f(g)$ is $F(g)/g'$ (in your example: $f(x)=\frac1x,F(x)=\ln(\vert x\vert),g(x)=x^2$). The probable misconception being that to differentiate this function we can just differentiate $F(g)$, whose derivative is $g'f(g)$, and that the $g'$ will cancel with $1/g'$. This is not so. You have a quotient $\frac{F(g)}{g'}$, so you need the quotient rule to differentiate, which gives

$$\frac{g'f(g)g'-F(g)g''}{g'^2}=f(g)-\frac{F(g)g''}{g'^2}$$

as the derivative. So you're stuck with the superfluous term on the right. In general, there is no easy rule to find antiderivatives of compositions of functions.

score 0 · Answer 3 · answered Jan 24 '22 at 15:59

If you have a function $f:(\infty,0)\cup(0,\infty)\rightarrow\mathbb{R}$ defined by $$f(x)=\begin{cases}\ln(-x)+A&x\lt0\\\ln(x)+B&x\gt0\end{cases}$$ for fixed $A,B\in\mathbb{R},$ then $$\forall{x\in(\infty,0)\cup(0,\infty)},\;f'(x)=\frac1{x}.$$ As such, we say that $$\int\frac1{x}\,\mathrm{d}x=\begin{cases}\ln(-x)+A&x\lt0\\\ln(x)+B&x\gt0\end{cases}.$$ On the other hand, if $$f(x)=\begin{cases}-\frac1{x}+A&x\lt0\\-\frac1{x}+B&x\gt0\end{cases}$$ for fixed $A,B\in\mathbb{R},$ then $$\forall{x\in(\infty,0)\cup(0,\infty)},\;f'(x)=\frac1{x^2}.$$ As such, we say that $$\int\frac1{x^2}\,\mathrm{d}x=\begin{cases}-\frac1{x}+A&x\lt0\\-\frac1{x}+B&x\gt0\end{cases}.$$ In fact, for the general case, what you will have is that $$\int{x^n}\,\mathrm{d}x=\begin{cases}\frac{x^{n+1}}{n+1}+A&n\neq-1,\,x\lt0\\\frac{x^{n+1}}{n+1}+B&n\neq-1,\,x\gt0\\\ln(-x)+A&n=-1,\,x\lt0\\\ln(x)+B&n=-1,\,x\gt0\end{cases}.$$ This is because there is no $A,n$ such that if $f(x)=Ax^n,$ then $$f'(x)=\frac1{x}.$$ On the other hand, notice that $$\frac{x^{n+1}}{n+1}=\frac{e^{\ln(x)(n+1)}}{n+1}=\frac{1+\ln(x)(n+1)+\sum_{m=2}^{\infty}\frac{\ln(x)^m(n+1)^m}{m!}}{n+1}$$ $$=\frac1{n+1}+\ln(x)+\ln(x)\sum_{m=1}^{\infty}\frac{\ln(x)^m(n+1)^m}{(m+1)!}.$$ What this implies is that $$\lim_{n\to-1}\frac{x^{n+1}}{n+1}-\frac1{n+1}=\ln(x),$$ and notice that $$\lim_{n\to-1}\frac{x^{n+1}}{n+1}-\frac1{n+1}=\lim_{n\to-1}\int_1^xt^n\,\mathrm{d}t,$$ so $$\lim_{n\to-1}\int_1^xt^n\,\mathrm{d}t=\ln(x).$$

Integrals of $1/x$ and $1/x^2$

3 Answers3