## anonymous 5 years ago integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^2-15)) should i do partial fraction 1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15) , or 1/((x^2*(x^2-15)) = A / x + B/x^2 + Cx+d / ( x^2 - 15)

1. anonymous

why don't you type in math. will help in viewing easily.

2. anonymous

the second one is the right way. How do you hope getting radicals will help the partial fraction??(Indeed, we use partial fractions for simplification!!!)

3. anonymous

hello

4. anonymous

whats wrong with partial fractions with radicals?

5. anonymous

i dont see any problem with the first way

6. anonymous

no problems but we avoid it as calculations get messy.

7. anonymous

my computer does the partial fractions integral 1/(x^2 (x^2-15)) = -1/(15x^2) + 1/(15(x^2-15))

8. anonymous

yah...that "seems" easy :) you can always take x^2=y during partial fracs.. then u get 1/(y(y-15)) which can be easily split into fracs.

9. anonymous

that doesnt disobey partial fractions rule

10. anonymous

i thought it was , it had to be linear factors and irreducibles

11. anonymous

like A/x + B/x^2 + C / (x-sqrt 15) + D/(x+sqrt 15)

12. anonymous

See...there are no so-called rules...at least not for me :) if we take radicals the math is correct but its unconventional

13. anonymous

cx+d is supposed to be only for irreducibles, i thought

14. anonymous

like integral 1/x(x^2+1) = A/x + bx+c / x^2 + 1

15. anonymous

so youre saying 1/x(x^2-1) = A/x + (bx+c) /( x^2 - 1)

16. anonymous

not only for irreducibles,

17. anonymous

im pretty sure thats what the theorem states

18. anonymous

maybe the theorem u saw said that we cx+d for irreducibles but it did not say that we cant use cx+d for non-irreducibles!!

19. anonymous

we use cx+d for irreducibles but we can also use it like (bx+c)/(x^2-1). Here the math is correct. But splitting like A/(x+1)+B/(x-1) is more advantageous, as it gives a good form whose integral is known.

20. anonymous

for example, this partial fractions fails 1/(x^2 (x-5) ) = A/x^2 + B/(x-5) , this fails

21. anonymous

oh because of the A/x + B/x^2

22. anonymous

yes

23. anonymous

hmmm

24. anonymous

ok so 1/[(x-a)^n (x^2 +bx +c)^n] = k1/(x-a) + k2/(x-a)^2 + ... kn/(x-a)^n + m1x+p1/(x^2 +bx +c) +...

25. anonymous

your upto kn is ok. but from m1,p1,...its getting complicated.

26. anonymous

yeah im running out of letters

27. anonymous

no. there will be cx+d term above and the denominator's power will increase from 1 to n

28. anonymous

1/[(x-a)^n (x^2 +bx +c)^n] = A1/(x-a) + A2/(x-a)^2 + ... An/(x-a)^n + B1x+C1/(x^2 +bx +c) + (B2x +C2)/(x^2 +bx + c)^2... +

29. anonymous

1/[(x-a)^n (x^2 +bx +c)^n] = A1/(x-a) + A2/(x-a)^2 + ... An/(x-a)^n + B1x+C1/(x^2 +bx +c) + (B2x +C2)/(x^2 +bx + c)^2... +(Bnx +Cn) / (x^2 +bx +c)^n

30. anonymous

your c1 and b1 terms are wrong

31. anonymous

why?

32. anonymous

there won't be any b1x ... also instead of c1 write it as b1x+c1

33. anonymous

1/[(x-a)^n (x^2 +bx +c)^n] = A1/(x-a) + A2/(x-a)^2 + ... An/(x-a)^n + (B1x+C1)/(x^2 +bx +c) + (B2x +C2)/(x^2 +bx + c)^2... +(Bnx +Cn) / (x^2 +bx +c)^n

34. anonymous

yups. thats correct....:)

35. anonymous

so thats fine?

36. anonymous

oh i left out the parenthesees

37. anonymous

ok so thats fine for partial fractions , and x^2 +bx + c doesnt necessarily have to be irreducible,

38. anonymous

see..make sure that you know that WE USE cx+d in numerator FOR IRREDUCIBLES (in other words polynomials have only complex roots). WE COULD use cx+d also above REDUCIBLES since it is mathematically correct but don't use it as one of the partial fractions is not in its lowest degrees (i.e. cx+d over a reducible is NOT recommended) But if such a use simplifies a problem at hand then why not use it? Understand me?

39. anonymous

40. anonymous

reducible into linear factors

41. anonymous

yups yups a QUADRATIC can only have cx+d above it. yes reducible into linear REAL factors. Nobody wants to have complex factors here!

42. anonymous

ok, so then we could do 1/(x^2 (x^2 - 15)) , we could reduce it

43. anonymous

im doing 1 / ( x ( x^2 + 2x + 1) ) the latter is reducible, but im not going to . = AX+b / ( x^2 + 2x + 1) + C / x

44. anonymous

Mathematically permissible. But not recommended.

45. anonymous

ok lets try this one 1/ ( x (x^2+2x+1)) integrate that using parfrac

46. anonymous

int 1/ ( x (x^2+2x+1))= (AX+b) / ( x^2 + 2x + 1) + C / x

47. anonymous

in part fracs.. its (1/x)+((-x-2)/(x^2+2x+1)) the first term gives log|x| and the second make substitution x^2+2x=y

48. anonymous

right, or we could have factored x^2 + 2x + 1

49. anonymous

yes...

50. anonymous

wait thats wrong

51. anonymous

whats that substitution ?

52. anonymous

yupss....the subs wrong...

53. anonymous

we can do it however

54. anonymous

oh by the way, partial fraction doesnt help us with say 1/ (x^(3/2) (x-15))

55. anonymous

no.. since the denominator is not a POLYNOMIAL

56. anonymous

, ahh

57. anonymous

both numer. and denom. must be polynomials

58. anonymous

wait is that a precondition?

59. anonymous

you could have 1/(x*(x^2-sqrt 3)) for example

60. anonymous

YES partial fractions can help ONLY TO RATIONAL FUNCTIONS i.e. both numeratoer and denominator polynomials with denominator not being the zero polynomial

61. anonymous

ok

62. anonymous

just clearing that up

63. anonymous

by the way what's your age?

64. anonymous

25

65. anonymous

why do i sound dumb

66. anonymous

hahha..joking? i hope u understood.. ask for doubts...

67. anonymous

i am 17

68. anonymous

oh

69. anonymous

jeez, i thought you were older. you are wise beyond your years :)

70. anonymous

are you a mathematician? or math-related worker? u named urself cantorset...

71. anonymous

yes im a math tutor

72. anonymous

for a living, i was stumped because the student didnt want to reduce a quadratic

73. anonymous

and i thought, hey the theorem says so and so

74. anonymous

it was reducible into irrational factors

75. anonymous

the question u asked me? - yes it can be solved both ways

76. anonymous

see this problem we just did its a lot easier if we reduce the quadratic

77. anonymous

integral (1/(x(x^2+2x+1))

78. anonymous

yes its lot easier if we reduce quad... but i argued just because i wanted to show that cx+d hold good for reducible quadratics also..

79. anonymous

right so i guess it depends on the problem

80. anonymous

well its good to know the power of this theorem

81. anonymous

which area of math are you specialized in?

82. anonymous

i tutor mostly calculus ,

83. anonymous

wow..thats awesome..i love calculus..but know little..struggling still with single variable calculus

84. anonymous

youre amazing :)

85. anonymous

well if you need help, give me a buzz

86. anonymous

ok..thank you.

87. anonymous

so yeah, this one is tougher to integrate , without reducing the quadratic

88. anonymous

integral (1/x)+((-x-2)/(x^2+2x+1))

89. anonymous

yups.

90. anonymous

so the original question integral 1/(x^2 (x^2-15))

91. anonymous

then reducing the quadratic is worse, lol

92. anonymous

x^2=y, integr=(1/15)log|(y-15)/y|+constant

93. anonymous

then you need 2x dx = dy

94. anonymous

oh its not so simple ,

95. anonymous

i.e. integral=$\frac{1}{15}\ln \left| \frac{x^2-15}{x^2} \right|+constant$

96. anonymous

when you make a substitution, you have to do dy = 2x dx

97. anonymous

SORRY!! out of my mind......u r right..

98. anonymous

then we have another problem

99. anonymous

we want dx by itself

100. anonymous

dy/(2x) = dx

101. anonymous

so we have dx = dy / (2sqrt y))

102. anonymous

ok, lets not do the substitution then

103. anonymous

$\frac{1}{15x}+\frac{1}{30\sqrt{15}}\ln \left| \frac{x-\sqrt{15}}{x+\sqrt{15}} \right|+constant$

104. anonymous

i think this is the answer.

105. anonymous

let me check

106. anonymous

i use maple

107. anonymous

1/(15*x)-(1/225)*sqrt(15)*arctanh((1/15)*x*sqrt(15))

108. anonymous

must be the same..

109. anonymous

oh i messed up

110. anonymous

so what did you use ?

111. anonymous

$\int\limits_{}^{}\frac{dx}{x^2-15}=\frac{1}{\sqrt{15}} \ln \left| \frac{x-\sqrt{15}}{x+\sqrt{15}} \right| +C$ i used this identity

112. anonymous

but theres an x^2 in front

113. anonymous

that is easy!!

114. anonymous

$\int\limits_{?}^{?}1/(x^2(x^2-15))$

115. anonymous

how do you do fraction in latex?

116. anonymous

frac{2}{3}=2/3

117. anonymous

1/(x^2(x^2-15)) = A/x + B/x^2 + (cx+d) / (x^2-15)

118. anonymous

interestingly enough, A = 0

119. anonymous

so its equal to integra (-1/(15x) + (1/(15(x^2-15))

120. anonymous

x^2=t substitution would save time. By the way, i hope the integrations are clear... i hav to leave...

121. anonymous

but you have to do a little work with the substitution

122. anonymous

yes..either way its correct.

123. anonymous

oh i think you got it

124. anonymous

so you fixed the substitution?

125. anonymous

yes..

126. anonymous

but thats weird

127. anonymous

ok x^2 = y, so dx = dy/(2x) = dy/(2sqrt y )

128. anonymous

so int 1/(x^2(x^2-15)) = int (1/( 2 sqrt y *y *(y-15))

129. anonymous

i used the substitution only for the partial fractions. not for the integrations. i am working with y in partial fracs but wreturning to x in integration

130. anonymous

oh

131. anonymous

it saves time in partial fracs..

132. anonymous

so you got integral 1/(x^2(x^2-15)) = integral (-1/(15x) + (1/(15(x^2-15)). and THEN did substitution

133. anonymous

no ..i used y=x^2 only for partial fracs..

134. anonymous

you cant do that, as far as i know

135. anonymous

That is allowed.

136. anonymous

oh then you substituted back before you integrated?

137. anonymous

working with a smaller degree variable in partial frac.. is very good idea. yah i subst. back before i integrated

138. anonymous

ok so you got integral 1/(x^2(x^2-15)) = integral (-1/(15y) + (1/(15(y-15)).

139. anonymous

Yes. that's right. i then changed y to x^2

140. anonymous

thats the same thing as just integral 1/(x^2(x^2-15)) = integral (-1/(15x^2) + (1/(15(x^2-15))

141. anonymous

oh i see, since its similiar to 1 / ( x (x-5))

142. anonymous

yes...but during partial fracs.. dealing with y is easier than dealing with x^2...its just the smae thing

143. anonymous

you are right

144. anonymous

right, but one has to be careful to switch back

145. anonymous

yes...i did a silly mistake back 15 mins. earlier

146. anonymous

youre basically using the identity 1/(u(u-15) = -1/15u + 1/15(u-15)

147. anonymous

yes

148. anonymous

ok

149. anonymous

substitution is way for complicated, i mean u substitution

150. anonymous

yes, but helps sometimes...avoid it when unnecessary...

151. anonymous

right, if we did u substitution, we would be able to do partial fractions anymore

152. anonymous

we would have u = x^2 , du = 2x dx, and dx = du/(2x) = du / (2 sqrt u) int ( 1/ (sqrt u * u * (u-15)) du

153. anonymous

yes..

154. anonymous

this is a very interesting problem, the pitfalls

155. anonymous

pitfall is like quicksand, lol

156. anonymous

so its better to say, i used the algebraic identity, blah blah

157. anonymous

one has to be lucky to get the right idea for integration

158. anonymous

so you dont confuse , lol

159. anonymous

i substituted into the algebraic identity, that would be best :)

160. anonymous

1/(u*(u-15)) = 1/ (15 u ) + ./.. etc

161. anonymous

whatever...everyone has a different way to approach problems ..:)

162. anonymous

thats fine

163. anonymous

its just when you say substitution and integration, most people would think you are doing u substitution

164. anonymous

really what it should be called is change of variable

165. anonymous

anyways, i see now why partial fractions fail on the complete u substitution

166. anonymous

its not a rational function, as you pointed out. thanks for your help. you've been great!

167. anonymous

yah...i should have shown the whole process but takes too much time in latex..then there would have been no confusions...by the way i m not too good in english

168. anonymous

169. anonymous

what country?

170. anonymous

where are u from? i am from india

171. anonymous

USA

172. anonymous

near new york

173. anonymous

k...celebrations there right? after osama was killed?

174. anonymous

i didnt celebrate

175. anonymous

he wasnt given a trial, that kind of thing

176. anonymous

i would like to see the evidence, and charges. but i guess it was stupid for him to take credit for it, if he really did take credit

177. anonymous

he could've escaped if he was allowed to live....nobody can predict him and his associates,....:)

178. anonymous

but he was a CIA asset, when he was fighting the russians

179. anonymous

im not crazy about our govt. too much deception

180. anonymous

i dont know if he is solely responsible for 911 , he might just be one piece of the puzzle

181. anonymous

whatever lets leave these to the concerned authorities... Hope u r ok with par fracs...bye

182. anonymous

take care