Indefinite Integral of (1/x^2)

[Liger20]

Homework Statement

Hello, first of all I would like to apologize for the fact that this question is extremely trivial compared to the other questions being asked. I have a improper integral problem, and the entire problem itself is not relevant, because I understand everything in it except for one thing. One step in the problem requires finding the indefinite integral of (1/x^2). The example in the book tells me that the answer is (-1/x), and it says that the answer is obtained by using the reverse power rule, but I just can’t see how they got that answer. I have a feeling that it is something very simple, and that I’ve forgotten some subtle detail.

Homework Equations

The Attempt at a Solution

Here’s how I tried to solve it:

(1/x^2) has an overall power of one, right? I increased the power of the whole thing by one, which is (1/x^2)^2, and I divided the whole thing by two, which is the same thing as multiplying by ½. So…

(½)(1/x^2)^2

You can already see that this is not going to give an answer of (-1/x). Could someone please help me with this?

 

[Avodyne]

$$\int x^n\,dx = {1\over n+1}x^{n+1} + C$$
In your case, $n=-2$.

 

[Liger20]

Hmmm...okay, I'll buy that. But why is n negative 2? If I'm raising the whole thing to the power of 2, why isn't n just positive 2? Thank you!

 

[kbaumen]

$$\frac{1}{x^n} = x^{-n}$$

A basic property of exponents.

 

[Cyosis]

Raising the entire expression to the power of 2 would give you $(1/x^2)^2=1/x^4=x^{-4}$.

 

[Mark44]

Hmmm...okay, I'll buy that. But why is n negative 2? If I'm raising the whole thing to the power of 2, why isn't n just positive 2? Thank you!

To expand on what kbaumen wrote, and relative to your problem,
$$\frac{1}{x^2} = x^{-2}$$