Christos
  • Christos
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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hartnn
  • hartnn
so where are you stuck ?
Christos
  • Christos
@hartnn
Christos
  • Christos
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 ! :(

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ganeshie8
  • ganeshie8
link is not opening for meh
Christos
  • Christos
http://screencast.com/t/WittgfBhztpV
Christos
  • Christos
just tell me what do I do
Christos
  • Christos
for the third one
Christos
  • Christos
after the long division , then what
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
*degree
Christos
  • Christos
*reading carefully*
hartnn
  • hartnn
yeah, thats the reason she did long division for 3rd
hartnn
  • hartnn
so, you must have something like x/x^2+1 ,right ??
hartnn
  • hartnn
hint : d/dx (x^2+1) = 2x
hartnn
  • hartnn
so take u = x^2+1 , du =...
ganeshie8
  • ganeshie8
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 :)
ganeshie8
  • ganeshie8
@hartnn ... plz continue if u have time :)
hartnn
  • hartnn
i have like 3 minutes :P
ganeshie8
  • ganeshie8
and im sure, u can make Christos understand half of calculus in less than 3 minutes ;)
hartnn
  • hartnn
probably, if she is sitting right in front of me ;)
Christos
  • Christos
she is intenting a new study method called parallel processing !!
Christos
  • Christos
I am just kidding
hartnn
  • hartnn
times up :P i need to rush b4 my bus leaves without me :P be back online when i reach home..
ganeshie8
  • ganeshie8
oh hartnn you're in office ha ... runn.. il see if i can do some justice in explaining partial fractions :)
ganeshie8
  • ganeshie8
\(\large \frac{1}{x^2+3x-4}\) factor the denominator \(\large \frac{1}{(x-1)(x+4)}\)
ganeshie8
  • ganeshie8
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
Christos
  • Christos
answer: 1/5ln|x-2| - 1/5 ln|x+4| + C
Christos
  • Christos
is it ?
ganeshie8
  • ganeshie8
thats right !
Christos
  • Christos
lets go 2nd
Christos
  • Christos
do I need to factor the denominator here ?
ganeshie8
  • ganeshie8
yes, factoring IS the first step
ganeshie8
  • ganeshie8
and ofc, the 0th step is long-division :)
Christos
  • Christos
A/x b/x+1 C/x+1
Christos
  • Christos
B/x-1 * * *
ganeshie8
  • ganeshie8
step1 : factor denominator \(\large \frac{1}{x(x^2-1)} \) \(\large \frac{1}{x(x-1)(x+1)} \)
Christos
  • Christos
A B C: http://screencast.com/t/a5HayvjoOi
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
yes looks good, solve A, B, C
Christos
  • Christos
I hope you are not leading me into a teaching trap at the current moment huh :D
Christos
  • Christos
ok ok I solve it now
ganeshie8
  • ganeshie8
whats teaching trap lol im no teacher... guess im more toward a professor who screws his students hmm lol
Christos
  • Christos
ok I can't solve this, whats the next step :D
Christos
  • Christos
if x = 1 denominator is undefined A.k.A me missing something
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
\(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) to sovle A : Multuply the whole equation wid \(x\), and plug \(x = 0\) \(\large \frac{1}{(x-1)(x+1)} = A + \frac{Bx}{x-1} + \frac{Cx}{x+1}\)
ganeshie8
  • ganeshie8
when u plug \(x = 0\), everything on right side cancels away, except \(A\)
Christos
  • Christos
-1
ganeshie8
  • ganeshie8
\(\large \frac{1}{(0-1)(0+1)} = A + 0 + 0\) solve A
ganeshie8
  • ganeshie8
yes, A = -1
Christos
  • Christos
yesyes
ganeshie8
  • ganeshie8
similarly find B and C
Christos
  • Christos
x = ?
ganeshie8
  • ganeshie8
\(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) to find B :- multiply the whole equation wid \((x-1)\) and plugin \(x = 1\) \(\large \frac{1}{x(x+1)} = \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x+1}\)
Christos
  • Christos
and then the main fraction is undefined..
ganeshie8
  • ganeshie8
how ?
ganeshie8
  • ganeshie8
\(\large \frac{1}{x(x+1)} = \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x+1}\) plugin x = 1 \(\large \frac{1}{1(1+1)} = 0 + B + 0\)
ganeshie8
  • ganeshie8
il let u solve C :)

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