Question

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)

Answer 1

so where are you stuck ?

Answer 2

@hartnn

Answer 3

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 ! :(

Answer 4

link is not opening for meh

Answer 5

http://screencast.com/t/WittgfBhztpV

Answer 6

just tell me what do I do

Answer 7

for the third one

Answer 8

after the long division , then what

Answer 9

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

Answer 10

*degree

Answer 11

*reading carefully*

Answer 12

yeah, thats the reason she did long division for 3rd

Answer 13

so, you must have something like
x/x^2+1 ,right ??

Answer 14

hint :
d/dx (x^2+1) = 2x

Answer 15

so take u = x^2+1 , du =...

Answer 16

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 :)

Answer 17

@hartnn ... plz continue if u have time :)

Answer 18

i have like 3 minutes :P

Answer 19

and im sure, u can make Christos understand half of calculus in less than 3 minutes ;)

Answer 20

probably, if she is sitting right in front of me ;)

Answer 21

she is intenting a new study method called parallel processing !!

Answer 22

I am just kidding

Answer 23

times up :P i need to rush b4 my bus leaves without me :P
be back online when i reach home..

Answer 24

oh hartnn you're in office ha ... runn.. il see if i can do some justice in explaining partial fractions :)

Answer 25

\(\large \frac{1}{x^2+3x-4}\)

factor the denominator

\(\large \frac{1}{(x-1)(x+4)}\)

Answer 26

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 27

answer: 1/5ln|x-2| - 1/5 ln|x+4| + C

Answer 28

is it ?

Answer 29

thats right !

Answer 30

lets go 2nd

Answer 31

do I need to factor the denominator here ?

Answer 32

yes, factoring IS the first step

Answer 33

and ofc, the 0th step is long-division :)

Answer 34

A/x b/x+1 C/x+1

Answer 35

B/x-1 * * *

Answer 36

step1 : factor denominator

\(\large \frac{1}{x(x^2-1)} \)

\(\large \frac{1}{x(x-1)(x+1)} \)

Answer 37

A B C: http://screencast.com/t/a5HayvjoOi

Answer 38

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

Answer 39

yes looks good, solve A, B, C

Answer 40

I hope you are not leading me into a teaching trap at the current moment huh :D

Answer 41

ok ok I solve it now

Answer 42

whats teaching trap lol im no teacher... guess im more toward a professor who screws his students hmm lol

Answer 43

ok I can't solve this, whats the next step :D

Answer 44

if x = 1 denominator is undefined A.k.A me missing something

Answer 45

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

Answer 46

\(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\)

to sovle A :
Multuply the whole equation wid \(x\), and plug \(x = 0\)
\(\large \frac{1}{(x-1)(x+1)} = A + \frac{Bx}{x-1} + \frac{Cx}{x+1}\)

Answer 47

when u plug \(x = 0\), everything on right side cancels away, except \(A\)

Answer 48

-1

Answer 49

\(\large \frac{1}{(0-1)(0+1)} = A + 0 + 0\)

solve A

Answer 50

yes, A = -1

Answer 51

yesyes

Answer 52

similarly find B and C

Answer 53

x = ?

Answer 54

\(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\)

to find B :-
multiply the whole equation wid \((x-1)\) and plugin \(x = 1\)
\(\large \frac{1}{x(x+1)} = \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x+1}\)

Answer 55

and then the main fraction is undefined..

Answer 56

how ?

Answer 57

\(\large \frac{1}{x(x+1)} = \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x+1}\)

plugin x = 1

\(\large \frac{1}{1(1+1)} = 0 + B + 0\)

Answer 58

il let u solve C :)