Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Please fellow open study members, solve with me . Finals countdown 3 hours :P https://www.dropbox.com/s/vwu945w28tgowlp/Screenshot%202014-01-15%2015.09.53.jpg (ANXIETY LEVEL OVER THE TOP)

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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

so where are you stuck ?
They are the only problems to which I don't have solutions , and I also have infinite series to cover , if you give me what I lack, it would be of a great help at the current situation ! :(

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

link is not opening for meh
http://screencast.com/t/WittgfBhztpV
just tell me what do I do
for the third one
after the long division , then what
before we dive in, two things to keep in mind :- 1) we can integrate ANY/EVERY rational function 2) to use partial fractions, degreem of numerator must always be LESS than degree of denominator
*degree
*reading carefully*
yeah, thats the reason she did long division for 3rd
so, you must have something like x/x^2+1 ,right ??
hint : d/dx (x^2+1) = 2x
so take u = x^2+1 , du =...
for first example, since degree of numerator < degree of denominator, next do below steps :- step1 : factor the denominator ive seen hartnn explaining partial fractions excellently before... so i give u hartnn :)
@hartnn ... plz continue if u have time :)
i have like 3 minutes :P
and im sure, u can make Christos understand half of calculus in less than 3 minutes ;)
probably, if she is sitting right in front of me ;)
she is intenting a new study method called parallel processing !!
I am just kidding
times up :P i need to rush b4 my bus leaves without me :P be back online when i reach home..
oh hartnn you're in office ha ... runn.. il see if i can do some justice in explaining partial fractions :)
\(\large \frac{1}{x^2+3x-4}\) factor the denominator \(\large \frac{1}{(x-1)(x+4)}\)
step2 : break left fraction into sum of fractions :- \(\large \frac{1}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4}\) find A and B
answer: 1/5ln|x-2| - 1/5 ln|x+4| + C
is it ?
thats right !
lets go 2nd
do I need to factor the denominator here ?
yes, factoring IS the first step
and ofc, the 0th step is long-division :)
A/x b/x+1 C/x+1
B/x-1 * * *
step1 : factor denominator \(\large \frac{1}{x(x^2-1)} \) \(\large \frac{1}{x(x-1)(x+1)} \)
A B C: http://screencast.com/t/a5HayvjoOi
step2 : wrote the fraction as sum of simple fractions \(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) solve A, B, C
yes looks good, solve A, B, C
I hope you are not leading me into a teaching trap at the current moment huh :D
ok ok I solve it now
whats teaching trap lol im no teacher... guess im more toward a professor who screws his students hmm lol
ok I can't solve this, whats the next step :D
if x = 1 denominator is undefined A.k.A me missing something
there is a small trick to solve this, it wont take any time to learn the trick. let me show u how to find A, B, C super fast
\(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) to sovle A : Multuply the whole equation wid \(x\), and plug \(x = 0\) \(\large \frac{1}{(x-1)(x+1)} = A + \frac{Bx}{x-1} + \frac{Cx}{x+1}\)
when u plug \(x = 0\), everything on right side cancels away, except \(A\)
-1
\(\large \frac{1}{(0-1)(0+1)} = A + 0 + 0\) solve A
yes, A = -1
yesyes
similarly find B and C
x = ?
\(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) to find B :- multiply the whole equation wid \((x-1)\) and plugin \(x = 1\) \(\large \frac{1}{x(x+1)} = \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x+1}\)
and then the main fraction is undefined..
how ?
\(\large \frac{1}{x(x+1)} = \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x+1}\) plugin x = 1 \(\large \frac{1}{1(1+1)} = 0 + B + 0\)
il let u solve C :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question