I need help with this and I need it fast!! (1 Viewer)

[Korn]

Just a maths question....



Find the indefinite integral:

S (x^2 + x + 3) / 3x^-5 .dx


I'm just stumbling heaps.... any help would be greatly appreciated!

This would be best answered if it were in the Maths forum in the school community
Put this in the Maths forum Please, Mods.

 

[Slidey]

S (x^2 + x + 3) / 3x^5 .dx

S (x^2 + x + 3) / 3x^5 .dx
= Integral of (1/x^3 + 1/x^4 + 3/x^5)/3
= Integral of (x^-3 + x^-4 + 3.x^-5)/3
= (-x^-2/2 -x^-3/3 -3x^-4/4)/3
= -(6/x^2 + 4/x^3 + 9/x^4)/36
= -(6x^2 + 4x + 9)/(36x^4)

 

[m_isk]

1/(2x^-2)

You know that anything to the power of a negative means you can do this: a^-2 = 1/a^2

therefore 1 / [ 1 / (2x^2) ]

this is just a division of fractions:

1/1 divided by 1/2x^2

= 1/1 . 2x^2

= 2x^2

actually, the way ur doing it assumes that it is (2x)^-2, which is not the case. 1/(2x^-2) is best done by saying that 1/(2x^-2) = 1 divided by 2x^-2. as you know, when you are dividing one thing by another, you can also multiply by its reciprocal, as nick said. therefore, 1 / 2x^-2= 1/ (2/x^2) but 1/(2/x^2) = 1*(x^2/2) = .5x^2, not 2x^2. It is much easier to see (and do) on paper.

 

[nick1048]

um no its not I answered that part oredy and slide is answering his next question...

 

[nick1048]

actually, the way ur doing it assumes that it is (2x)^-2, which is not the case. 1/(2x^-2) is best done by saying that 1/(2x^-2) = 1 divided by 2x^-2. as you know, when you are dividing one thing by another, you can also multiply by its reciprocal, as nick said. therefore, 1 / 2x^-2= 1/ (2/x^2) but 1/(2/x^2) = 1*(x^2/2) = .5x^2, not 2x^2. It is much easier to see (and do) on paper.

I dont understand your working out, I dont think it's right.

 

[m_isk]

1/3x^-5 = 1/1/3x^5

= 1/1 divided by 1/3x^5

= 1/1 . 3x^5

= 3x^5

believe me im right. the same flaw is present in your post on the previous page. In 1/3x^-5, it is only the x which is to the power of -5, not the 3. So, 1/3x^-5 = 1 divided by 3x^-5, which equals 1/(3/x^5) which equals (x^5)/3. Just do it on paper, and let your first step be changing the fraction notation to division...

 

[nick1048]

ahhhhh yea I see ur quite right there, my appologies.

 

[HellVeN]

Wtf.. I think you're like all wrong, even that website.

I don't think I've made any mistakes here, it was a pretty easy integral, my answer is:

Find the indefinite integral:
S (x^2 + x + 3) / 3x^-5 .dx



= 1/3 (((x^8)/8)+((x^7)/7)+((x^6)/2)) + C


I must be right!

 

[nick1048]

yeah the website just multiplies thru by your 1/3 lol... which is a factorisation step... so technically we're all right... my one just doesn't bring out the 1/3 constant... *sigh*

 

[HellVeN]

Hehe, well anyway here's a clear working out for the original post creator. I wanted to test out my scanner anyway.

 

[Slidey]

i copied the question incorrectly.... the same thing would apply with x^2 cancelling in this instance.....

S (x^2 + x + 3) / 3x^5 .dx (note the 5 is positive)

wouldn't it?

To answer your question: Yes. Divide through by x^2 and integrate each term.

 

[HellVeN]

Oh well.....................