Partial integration question

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Partial integration question

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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hold on i didnt know you couldnt post pic from iphone jsut 1 sec
draw your question, maybe...
i have done a lot of the steps so just 1 sec

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but heres the question |dw:1359962006954:dw|
that will be pretty exhausting..
x^4 (x+3) A/x+B/x^2+C/x^3+D/x^4 +E/(x+3)
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do you see my steps ill try posting another pic
yes, D=1, E=4 are correct.
yea i got that far phew but how do i solve for the A B and C after that?
you'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...
can you show me what you mean i dont get what you mean by that
should really have listened in my algebra classes sorry :(
\(Ax^3(x+3)+Bx^2(x+3)+Cx^3(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^2-8x+3\) put x = 1, to get 1 equation in A,B, C
it should be Cx (x+3)
another method is comparing the co-efficients to get 3 equations... yes, typo...it should be Cx(x+3)
like co-efficient of x^4 will be A+4 and no right side, its 9 so, A+4= 9 gives A=5
i think you should proceed with comparing the co-efficients..
so if you plug in x = 1 you get 4A + 4B+ 4C = 16
i think comapring the co-efficients will give you answer much faster, if you have understood it.
sorry internet disconnected agian
co-efficient of x^3 = 3A+B = 15+B on left on right its 17 so, 15+B = 17 gives B=3
forget about substitution...
ok 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
where did you get the|dw:1359963485018:dw|
it should be B+3A =17
then i also guess i dont get how you got the A + 4 either because my logic was wrong
|dw:1359963578884:dw|
not getting the |dw:1359963731639:dw|
\(Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^2-8x+3 \\ ~ \\Ax^4+3Ax^3 +Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4= \\ =9x^4+17x^3+3x^2-8x+3\)
do you have a formula or something that tells us that or how does that work?
\(Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^2-8x+3 \\ ~ \\Ax^4+3Ax^3 +Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4= \\ =9x^4+17x^3+3x^2-8x+3 \\ ~ \\ (A+4)x^4+(3A+B)x^3+....=...\)
comapring the c-efficiens on left and right, A+4=9 3A+B=17 did you get this ?
its not showing it in nice formatting so im writing it down just 1 sec
Ax^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+...
i just combined like terms on left..

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