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