anti diff question

[recovered]

hey guys reaally quick this simple "types" of questions

so anti diff of linear divided by a linear
for ex anti diff 2x+1/x-5

also lower power polynomial / higher power
ex x^3 +x+1/ x^4 +9

thnk u so much guys

[xZero]

Split it up into partial fractions

[b^3]

if you still want the solution or just to split it into partial fractions here it is. first time with LaTeX so excuse mistakes

so now anti diff it


I hope I typed the LaTeX code right

[yawho]

while on the topic of partial fractions, does anyone know why
(2x^5-8x^3+6x)/(x^6-6x^4+12x^2-8) = 2x(x-1)(x+1)(x^2 - 3)/(x^2 - 2)^3 = Ax/(x^2 - 2) + Bx/(x^2 - 2)^3 ?

[soopertaco]

or if that algebra trick didn't occur to you, you could've just long divided haha

[yawho]

I thought you long divide when the degree of the numerator is equal/higher than the degree of the denominator. How does one know that (2x^5-8x^3+6x)/(x^6-6x^4+12x^2-8) can be written as partial fractions in the form Ax/(x^2 - 2) + Bx/(x^2 - 2)^3 ?

[soopertaco]

yeah if you long divide you'll get b^3's result in one step

as for your other question, is that the original function you've given us?

[yawho]

yes

[b^3]

For the other question the following method will work to get it into the form ax/x^2-2 +bx/(x^2-2)^3, butim not sure if it will work by just long division as there is no complete remainder, anyway if you want it read below.





so the two middle terms add to zero

well there may be a shorter method but this works, just takes abit more time, anyway hope it helps

[yawho]

my tutor's way:
(2x^5-8x^3+6x)/(x^6-6x^4+12x^2-8)
=[2x(x^4-4x^2+4)-2x]/(x^6-6x^4+12x^2-8)
=[2x(x^2-2)^2-2x]/(x^2-2)^3
=2x/(x^2-2) - 2x/(x^2-2)^3

[b^3]

yeh thats simpler, wish i had of saw the shortcut before