Find the antiderivative

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Homework Statement



find the antiderivative of 1/(x^2)


The Attempt at a Solution



I'm pretty sure you just find the antiderivatives of the numerator and the denominator.

the antiderivative of 1 is x.
the antiderivative of x^2 is (x^3)/3
mutliply the numerator by the inverse of the deominator and you get 3/(x^2)

The book says the answer is -1/x
 

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  • #2
Ray Vickson
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Homework Statement



find the antiderivative of 1/(x^2)


The Attempt at a Solution



I'm pretty sure you just find the antiderivatives of the numerator and the denominator.

the antiderivative of 1 is x.
the antiderivative of x^2 is (x^3)/3
mutliply the numerator by the inverse of the deominator and you get 3/(x^2)

The book says the answer is -1/x

Have you ever seen the result ∫ x^n dx = x^(n+1)/(n+1) + C? It does not hold for n = -1 because that would involve division by zero on the right-hand-side, but otherwise n is unrestricted.

Please, please, please *get rid of the idea forever* that you can do the integral by integrating the numerator and denominator separately: that does not work! For example, integrate x, to get x^2/2. Now write x as x^2 / x and try your method of integrating the numerator and denominator separately---you will get the wrong answer.

RGV
 
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... (posted an idea that was false, subsequently deleted it)
 
  • #4
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Then how do I find the antiderivative of a fraction?

Here's another problem

1. 2/3 sec^2 x/3

2. (2/3 tan x/3)/(x/3)

3. (2/3 tan x/3) * (3/x)

4. 2/3 * 3/x (tan x/3)

5. 2/x tan x/3

the book says the answer is

2 tan (x/3)
 
  • #5
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Ok, I saw from another website, that to find the antiderivative of a fraction you have to convert the fraction into a number with an exponent.

so -1/x = -x^-1

apply (x^n+1)/(n+1)

that comes to

-(x^0)/0
 
  • #6
SammyS
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Then how do I find the antiderivative of a fraction?

Here's another problem

1. 2/3 sec^2 x/3

2. (2/3 tan x/3)/(x/3)

3. (2/3 tan x/3) * (3/x)

4. 2/3 * 3/x (tan x/3)

5. 2/x tan x/3

the book says the answer is

2 tan (x/3)
Assuming that you mean to find the anti-derivative of (2/3) sec2 (x/3) :

Rather than trying to follow those steps, which I can't make sense of ... Let's see if the book's answer is correct.

If 2 tan(x/3) is an anti-derivative of (2/3) sec2 (x/3), then the derivative of 2 tan(x/3), with respect to x, should be (2/3) sec2 (x/3) .

[itex]\displaystyle \frac{d}{dx}\tan(x)=\sec^2(x)[/itex]

So that: [itex]\displaystyle \frac{d}{dx}\left(2\,\tan\left(\frac{x}{3}\right) \right)=2\,\sec^2\left(\frac{x}{3}\right)\cdot \frac{1}{3}=\frac{2}{3}\, \sec^2\left(\frac{x}{3}\right)\,.[/itex]

The book's answer is correct.
 
  • #7
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Knowing the answer than finding the derivative can be done, but I still don't see how to find the answer starting from the antiderivative.

I got the answer to 1/x^-2 now.
 
  • #8
SammyS
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Ok, I saw from another website, that to find the antiderivative of a fraction you have to convert the fraction into a number with an exponent.

so -1/x = -x^-1

apply (x^n+1)/(n+1)

that comes to

-(x^0)/0
The anti derivative of xn=xn+1/(n=1), except if n = -1.

Why that exception?
It's because x0=1, and the derivative of 1 is 0, not x-1.​

The derivative of x-1 = ln(|x|)
 
  • #9
SammyS
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Knowing the answer then finding the derivative can be done, but I still don't see how to find the answer starting from the antiderivative.

I got the answer to 1/x^-2 now.
The anti-derivative of 1/x-2 is easy, because 1/x-2 = x2.

Earlier, you wanted the anti-derivative of 1/x2, which you found to be -1/x . That's easy to get if you write 1/x2 as x-2 .

Details ... details.
 
  • #10
Ray Vickson
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Did you fail to read my first post, which specifically told you this result fails for n = -1?

RGV
 

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