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Math Help Available

tomspug

tomspug

Absorbant
Hello, saw this was a new section of RF and just wanted to offer help in the mathematical areas! Just let me know if you need help...
 
darkendless

darkendless

Guardian of Asgaard
tomspug said:
Hello, saw this was a new section of RF and just wanted to offer help in the mathematical areas! Just let me know if you need help...
Click to expand...

Hello,

its 3 days before i can see my lecturer again, i thought i'd ask here. How good are you at integration?

i have a problem: the integral of 2x^2 - x +4 / x^3 +4x

I know the jist of it, im just not sure how to break the cubic function or factorise it in order to have nice looking fractions to solve.

Any help would be appreciated :)
 
Worshipper

Worshipper

Active Member
It's been years since I've done this, so I hope my help really is help!

Looking at that denominator, though, it seems to me that you might be able to factor out an x, leaving you with x(x^2 + 4). That might then be a good candidate for splitting up into partial fractions, especially since the quadratic factor there is a sum of two squares, which I think gets into certain trigonometric identities.

As a review of partial fractions, to split this rational expression into partial fractions, you solve the following equation for A, B, and C:

(2x^2 - x +4) / (x^3 +4x) = (A / x) + ((Bx + C) / (x^2 + 4))

Note the linear numerator to go with the irreducible quadratic denominator.

Does that help at all?
 
darkendless

darkendless

Guardian of Asgaard
Worshipper said:
It's been years since I've done this, so I hope my help really is help!

Looking at that denominator, though, it seems to me that you might be able to factor out an x, leaving you with x(x^2 + 4). That might then be a good candidate for splitting up into partial fractions, especially since the quadratic factor there is a sum of two squares, which I think gets into certain trigonometric identities.

As a review of partial fractions, to split this rational expression into partial fractions, you solve the following equation for A, B, and C:

(2x^2 - x +4) / (x^3 +4x) = (A / x) + ((Bx + C) / (x^2 + 4))

Note the linear numerator to go with the irreducible quadratic denominator.

Does that help at all?
Click to expand...

Excellent thankyou that was the exact clarification i needed. I thought A/X didnt seem right.

Thanks for your help, much appreciated :)
 
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