integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^2-15)) should i do partial fraction 1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15) , or 1/((x^2*(x^2-15)) = A / x + B/x^2 + Cx+d / ( x^2 - 15)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^2-15)) should i do partial fraction 1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15) , or 1/((x^2*(x^2-15)) = A / x + B/x^2 + Cx+d / ( x^2 - 15)

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

oy... if you hit the enter key it spreads this out to where it is more readable.... just a thought
the x^2 - 15 is reducible , but im not sure if youre supposed to use that theorem
enter key?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

the key on the keyboard that makes this happen ;)
you want me to use the enter key more often?
yes, difference of squares that part and you should have 3 sets to figure out
but you get radicals
radicals are fine.... right?
are they ? how come the other way works just as well?
which other way can you decompose the fraction?
1/((x^2*(x^2-15)) = A / x + B/x^2 + (Cx+d) / ( x^2 - 15)
technically the quadratic is reducible
it could be a fluke if it does work out....
or it could be a possibility for solutions that i am unaware of as yet :)
no it isnt a fluke, in general one need not reduce a quadratic
in general you should, to account for multiples i think
if it aint got multiples, then you should be fine i spose
right?
yes it usually makes it easier to reduce the quadratic, except in the case where there are radical linear factors
it can have multiples, theres no problem as far as i can see
without doing a ton of them to generalize it, id have to throw it in the hmmmmm box lol
say integral ( 1 / (x(x+1)^2) = A/x + (Bx+C) / (x+1)^2
its a little weird, might be easier if i wrote 1/( x ( x^2+2x+1)) = A/x + (Bx+c) / (x^2 + 2x + 1)
sometimes it is difficult to factor a quadratic, right?
ok , i should get into more accurate terms. sometimes it is reducible over reals, not rationals
factoring a quadratic is fairly easy; might get complicated solutions, but farily easy to do
yes but doing partial fractions with irrational roots is tough
and i have heard that you can use the complex solutions as well with no ill effects on the decomp
ok , so youre so confident, try to do parfractions on 1 / ( x^2(x^2-15) , give it a go
i was stuck
1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15)
confident? lol ; just stubborn
it can even get stranger, this is an easy irrational solution problem, you can have say A/ ( x - sqrt 3/2 ) for instance
this is why calculus in textbooks is generally contrived , or idealized problems. in the real world you have these irrational roots, etc
theorem time, one sec
1 a b c d ----------- = --- + ----- + --------- + --------- x^2(x^2-15) x x^2 x-sqrt(15) x+sqrt(15) is the 'proper' set up right?
yes looks good
Personally, I'd leave the last fraction's denominator without doing difference of two squares.
1 / [(x-a)^n (x^2+bx+c)^n] = A1/(x-a) + A2/ ( x-a)^2 + ... An/(x-a)^n + (B1x+C1)/ (x^2+bx+c) + (B2x + C2) / (x^2+bx+c)^2 + ... + (Bn x + Cn ) / (x^2+bx+c)^n
so the question is, do the quadratics have to be irreducible, i dont think so
1 a b c d ----------- = --- + ----- + --------- + --------- x^2(x^2-15) x x^2 x-sqrt(15) x+sqrt(15) a(x^2)(x-sqrt(15))(x+sqrt(15))) b(x)(x-sqrt(15))(x+sqrt(15))) c(x)( x^2)(x+sqrt(15)) d(x)(x^2)(x-sqrt(15))
MOST calculus books and websites say the quadratic is irreducible though
irreducible over the reals i imagine. wikipedia is better, they say irreducible over the field
whatever field you have in mine
For this particular problem: \[\frac{1}{y(y-15)} \equiv \frac{1}{15(y-15)} - \frac{1}{15y}\] \[\text{Let } y = x^2\] ...
newton, be careful though
if you are substituting a variable, you will need to do 2x dx = dy
so you might want to just change it back. thats an identity
i should write a book on calculus pitfalls, since i keep finding them :)
newtons, so just change y back to x^2
Sorry, in case it wasn't clear, it leads to: \[\frac{1}{x^2(x^2-15)} \equiv \frac{1}{15(x^2-15)} + \frac{1}{15x^2} \] I would integrate this.
otherwise you have 2x dx = dy , dx = dy / ( 2x ) = dy / ( 2 sqrt y )
correct, i thought you were doing a change of variables
1 = a(x^2)(x-sqrt(15))(x+sqrt(15))) + b(x)(x-sqrt(15))(x+sqrt(15))) + c(x)( x^2)(x+sqrt(15)) +d(x)(x^2)(x-sqrt(15)) when x = sqrt(15) we get 1 = c(sqrt(15))(15)(2sqrt(15)) 1 = c(15)(2)(15) 1 = c(225)(2) = c450 c = 1/450
amistre, you get medal for persistence :)
but its much easier if we dont split the quadratic
I was only doing that to show that you don't need a 1/x fraction and 1/x^2 , as x^2 is the only term. I wasn't integrating with a different variable.
yay!! lol :) can I stop now lol
ok
wait that doesnt work though
I mixed up a minus and plus sign ¬_¬
its incorrect to do 1 / ( x^2 (x^2-15 ) = A / x^2 + b / (x^2-15) , thats incorrect
you wont be able to find A and B
I just did?
you used a simpler case though
Multiply it out, plz.
1 = A(x^2 - 15) + B x^2
, 1 = Ax^2 + bx^2 - 15
(the + the last fraction should be minus, as in the first time, typo)
1 = Ax^2 + Bx62 - 15 A
hmmmm,
but the book says it should be A/ x + B/x^2 + C x+d / (x^2 - 15)
well this approach works as well
You can do it if you want, but you'll find A = 0, so it doesn't really matter.
right, so come the theorem doesnt work in these websites
as far as repeated roots go
for instance you didnt use CX+D, i think its because of that y = x^2 business,
There are only terms in x^2 in this case, it wouldn't always work. The general method is probably more ... general.
ahh
so my question, my original question, hmmmm, do the quadratics have to be irreducible in the general case
ok for instance, 1 / (( x-2)^2 ( x^2-15))
1 / (( x-2)^2 ( x^2-15)) = A / (x-2) + b / (x-2)^2 + (cx+d) / (x^2 - 15 ) ?
I'd say it's probably easier to leave it as the irreducible (without using difference of two squares) quadratic with Cx + d. I guess you could split up this factor, but it's unnecessary (as far as I know)
Yes, that.
but its not technically irreducible
i can even go so far as to do this
1 / (( x-2)^2 ( x^2-15)) = (Ax+b) / (x-2)^2 + (cx+d) / (x^2 - 15 )
no chat room
sorry, not for you
so this whole irreducible business is unnecessary
1 = a*x^3-15*a*x+b*x^2-15*b+c*x^3-4*c*x^2+4*c*x+d*x^2-4*d*x+4*d
ok im going to solve an easier one
int 1/ ((x-2)(x^2-15))
1/ ((x-2)(x^2-15)) = A/ (x-2) + (Bx+c) / ( x^ 2 - 15)

Not the answer you are looking for?

Search for more explanations.

Ask your own question