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so where are you stuck ?

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

*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 =...

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..

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

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

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

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

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

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 = ?

and then the main fraction is undefined..

how ?

il let u solve C :)