Interpretation of power rule for integration applied to 1/x

[daviddoria]

We all know
$\int \frac{1}{x} dx = ln(x) + c$

but if you try to apply the power rule for integration:
$\int x^n dx = \frac{x^{n+1}}{n+1} + c$

you get
$\int x^{-1} dx = \frac{x^0}{0}$

What can you learn from this/what does this mean?

David

 

[losiu99]

Well, I suppose we could just write
$$\int x^{-1+h} \mbox{d}x=\frac{x^h}{h}-C$$
and take C = 1/h. Then
$$\int x^{-1+h} \mbox{d}x=\frac{x^h-1}{h}\rightarrow\log x$$
(the limit as h -> 0). It's just a random thing that popped to my mind, but I liked it and decided to share.

 

[HallsofIvy]

We all know
$\int \frac{1}{x} dx = ln(x) + c$

but if you try to apply the power rule for integration:
$\int x^n dx = \frac{x^{n+1}}{n+1} + c$

you get
$\int x^{-1} dx = \frac{x^0}{0}$

What can you learn from this/what does this mean?

David

Since you cannot divide by 0, you learn that the power rule- $\int x^n dx= x^{n+1}/(n+1)+ C$- does not apply to the case n= -1. And I think you will find that stated in any textbook that gives the power rule for integration.