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sat_chen
 3 years ago
Partial integration question
sat_chen
 3 years ago
Partial integration question

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sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0hold on i didnt know you couldnt post pic from iphone jsut 1 sec

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1draw your question, maybe...

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0i have done a lot of the steps so just 1 sec

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0but heres the question dw:1359962006954:dw

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1that will be pretty exhausting..

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1x^4 (x+3) A/x+B/x^2+C/x^3+D/x^4 +E/(x+3)

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0do you see my steps ill try posting another pic

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1yes, D=1, E=4 are correct.

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0yea i got that far phew but how do i solve for the A B and C after that?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1you'll need to put 3 different values of x say x=1,1,2 then you get 3 equations in A,B, C with 3 unknowns...

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0can you show me what you mean i dont get what you mean by that

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0should really have listened in my algebra classes sorry :(

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1\(Ax^3(x+3)+Bx^2(x+3)+Cx^3(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^28x+3\) put x = 1, to get 1 equation in A,B, C

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1another method is comparing the coefficients to get 3 equations... yes, typo...it should be Cx(x+3)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1like coefficient of x^4 will be A+4 and no right side, its 9 so, A+4= 9 gives A=5

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1i think you should proceed with comparing the coefficients..

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0so if you plug in x = 1 you get 4A + 4B+ 4C = 16

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1i think comapring the coefficients will give you answer much faster, if you have understood it.

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0sorry internet disconnected agian

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1coefficient of x^3 = 3A+B = 15+B on left on right its 17 so, 15+B = 17 gives B=3

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1forget about substitution...

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0ok so what you are saying is that for A you A + 4 = 9 then for B would you B + 3 =17 then for C would you C + 2 =3

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0where did you get thedw:1359963485018:dw

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0then i also guess i dont get how you got the A + 4 either because my logic was wrong

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0not getting the dw:1359963731639:dw

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1\(Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^28x+3 \\ ~ \\Ax^4+3Ax^3 +Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4= \\ =9x^4+17x^3+3x^28x+3\)

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0do you have a formula or something that tells us that or how does that work?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1\(Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^28x+3 \\ ~ \\Ax^4+3Ax^3 +Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4= \\ =9x^4+17x^3+3x^28x+3 \\ ~ \\ (A+4)x^4+(3A+B)x^3+....=...\)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1comapring the cefficiens on left and right, A+4=9 3A+B=17 did you get this ?

sat_chen
 3 years ago
Best ResponseYou've already chosen the best response.0its not showing it in nice formatting so im writing it down just 1 sec

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4=9x^4+17x^3+3x^2−8x+3 Ax^4+3Ax^3+Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4=9x^4+17x^3+3x^2−8x+3 (A+4)x^4+(3A+B)x^3+....=9x^4+17x^3+...

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1i just combined like terms on left..
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