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Please fellow open study members, solve with me . Finals countdown 3 hours :P (ANXIETY LEVEL OVER THE TOP)

I got my questions answered at in under 10 minutes. Go to now for free help!
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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 ! :(

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Other answers:

link is not opening for meh
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
*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:
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 \(A\)
\(\large \frac{1}{(0-1)(0+1)} = A + 0 + 0\) solve A
yes, A = -1
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 :)

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