## Christos one year ago 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)

1. hartnn

so where are you stuck ?

2. Christos

@hartnn

3. 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 ! :(

4. ganeshie8

link is not opening for meh

5. Christos
6. Christos

just tell me what do I do

7. Christos

for the third one

8. Christos

after the long division , then what

9. 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

10. ganeshie8

*degree

11. Christos

12. hartnn

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

13. hartnn

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

14. hartnn

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

15. hartnn

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

16. 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 :)

17. ganeshie8

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

18. hartnn

i have like 3 minutes :P

19. ganeshie8

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

20. hartnn

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

21. Christos

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

22. Christos

I am just kidding

23. hartnn

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

24. ganeshie8

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

25. ganeshie8

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

26. 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

27. Christos

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

28. Christos

is it ?

29. ganeshie8

thats right !

30. Christos

lets go 2nd

31. Christos

do I need to factor the denominator here ?

32. ganeshie8

yes, factoring IS the first step

33. ganeshie8

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

34. Christos

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

35. Christos

B/x-1 * * *

36. ganeshie8

step1 : factor denominator $$\large \frac{1}{x(x^2-1)}$$ $$\large \frac{1}{x(x-1)(x+1)}$$

37. Christos
38. 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

39. ganeshie8

yes looks good, solve A, B, C

40. Christos

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

41. Christos

ok ok I solve it now

42. ganeshie8

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

43. Christos

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

44. Christos

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

45. 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

46. 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 $$x = 0$$ $$\large \frac{1}{(x-1)(x+1)} = A + \frac{Bx}{x-1} + \frac{Cx}{x+1}$$

47. ganeshie8

when u plug $$x = 0$$, everything on right side cancels away, except $$A$$

48. Christos

-1

49. ganeshie8

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

50. ganeshie8

yes, A = -1

51. Christos

yesyes

52. ganeshie8

similarly find B and C

53. Christos

x = ?

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

55. Christos

and then the main fraction is undefined..

56. ganeshie8

how ?

57. ganeshie8

$$\large \frac{1}{x(x+1)} = \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x+1}$$ plugin x = 1 $$\large \frac{1}{1(1+1)} = 0 + B + 0$$

58. ganeshie8

il let u solve C :)