# Integrating a polynomial

1. Jan 13, 2007

### snowJT

I Just want to know if this is how this should be done..

$$\frac{dy}{dx} = \frac{3y^2}{x}$$

$$dy = \frac{3y^2}{x}dx$$

$$dy = 3^2 x^-^1dx$$

$$\frac {1}{3} \int \frac {dy}{y^2} = \int x^-^1dx$$

$$\frac {1}{3} ln| y^2| = ln | x | + C$$

because I forget how to integrate the polynomial...

2. Jan 13, 2007

### cristo

Staff Emeritus
The right hand side is correct. Look again at $$\int y^{-2}dy$$ How do you integrate this?

3. Jan 13, 2007

### AlephZero

$$\frac {1}{3} \int \frac {dy}{y^2} = \int x^-^1dx$$

That's OK, but the integral of 1/y^2 isn't ln(y^2).

Look up how to integrate polynomials, if you forgot. Or if you know how to differentiate polynominals, work it out from the fact that integration is "anti-differentiation".

4. Jan 13, 2007

### snowJT

its not $$\frac{1}{3y}$$ is it?

5. Jan 13, 2007

### cristo

Staff Emeritus
Close; it's -1/(3y), since you must divide by the power of y(which is -1)

6. Jan 13, 2007

### snowJT

oops.. thanks I was thinking that was how you did it originally, I was just confused and did it that other way and forgot ket things... thanks

EDIT: Hmmm I also need to then solve for Y

$$y^-^1 = \frac {-ln|x|}{3} - \frac {C}{3}$$

$$y = \frac{-3}{ln|x|} - \frac{3}{C}$$

Last edited: Jan 13, 2007
7. Jan 13, 2007

### JJ420

that looks wrong to me

8. Jan 13, 2007

### JJ420

not even close i dont think

9. Jan 13, 2007

### JJ420

$$y=\frac{ln x + C}^-1{3}$$

Last edited by a moderator: Jan 13, 2007
10. Jan 13, 2007

### cristo

Staff Emeritus
$$y^{-1} = \frac {-ln|x|}{3} - \frac{C}{3}=\frac{-(ln|x|+C)}{3}$$

$$y=\frac{-3}{(ln|x|+C)}$$

Can you follow what I've done?

JJ420; we are here to help, not criticise!

11. Jan 13, 2007

### JJ420

$$y=\frac{ln x + C}^-1{3}$$
thats what i got but im dumb so dont trust me

12. Jan 13, 2007

### JJ420

oops that -13 is supposed to be an exponent of the numerator (-1) and 3 in the denominator

13. Jan 13, 2007

### cristo

Staff Emeritus
$$\frac{(ln x + C)^{-1}}{3}=\frac{1}{3(ln x + C)}$$ The 3 should definitely be in the numerator, and we need a - sign too. See my post:

Last edited: Jan 13, 2007
14. Jan 13, 2007

### snowJT

no I can't figure it out, common denominator?

15. Jan 13, 2007

### snowJT

oh.. yes I see

why is C positive and not negative

Last edited: Jan 13, 2007
16. Jan 13, 2007

### cristo

Staff Emeritus
Yea you just write ln|x| and C as one fraction with denominator 3. Then take the reciprocal of both sides (since the LHS is 1/y) to obtain an expression in terms of y.

Well, yes. In order to take the reciprocal of both sides, there must only be one fraction on each side.

17. Jan 13, 2007

### JJ420

looking back to the origonal i don't see how there is a negative

18. Jan 13, 2007

### cristo

Staff Emeritus
The minus sign comes about when integrating; integral(y-2)=-y-1 (+C)

19. Jan 13, 2007

### sara_87

snowJT what does C mean?

it means any constant...it can be a minus but since we don't know we just write + C

20. Jan 13, 2007

### snowJT

figured so, there were some examples I just found in my book