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anonymous

  • one year ago

Calculate the following integral.

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  1. anonymous
    • one year ago
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  2. freckles
    • one year ago
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    well we could do partial fractions again

  3. freckles
    • one year ago
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    \[x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)\] there is only 3 factors to worry about on the denominator

  4. freckles
    • one year ago
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    so it won't be too terrible

  5. freckles
    • one year ago
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    \[\frac{x^3-2}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1} \\ x^3-2=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+1)\] so multiply out and reorder then compare both sides

  6. freckles
    • one year ago
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    Do you want to try to find A,B,C, and D ?

  7. anonymous
    • one year ago
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    i got (A+B+C)X^3 + (A-B)x^2 + (A+B-C+D)x +(A-B-D) = x^3−2

  8. anonymous
    • one year ago
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    SO

  9. anonymous
    • one year ago
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    (A+B+C)=1 A+B=0 (A+B-C+D)=0 (A-B-D)=-2

  10. Zarkon
    • one year ago
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    you are working too hard

  11. anonymous
    • one year ago
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    :\ I only know how to expand and regroup

  12. freckles
    • one year ago
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    \[A(x^3+x^2+x+1)+B(x^3-x^2+x-1)+C(x^3-x)+D(x^2-1)=x^3-2 \\ x^3(A+B+C)+x^2(A-B+D)+x(A+B-C)+(A-B-D)=x^3-2 \\ A+B+C=1 \\ A-B+D=0 \\ A+B-C=0 \\ A-B-D=-2\]

  13. Zarkon
    • one year ago
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    \[\frac{x^3-2}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1} \\ x^3-2=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+1)\] let x=1

  14. Zarkon
    • one year ago
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    \[1^3-2=A(1+1)(1^2+1)+B(1-1)(1^2+1)+(C1+D)(1-1)(1+1)\] \[1-2=A(1+1)(1+1)\] A=-1/4

  15. Zarkon
    • one year ago
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    let x=-1 and you can get B let x=0 you can get d let x=i and you can get C

  16. freckles
    • one year ago
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    if you were having trouble solving the system, this may help I would add the first two equations together and the second two equations together: giving you: \[2A+(C+D)=1 \\ 2A-(C+D)=-2 \\ \text{ solving for } A \text{ by adding these two equations together } \\ 4A=-1 \\ A=\frac{-1}{4} \\ \text{ now see the equations that just have } A,B,C \text{ we can solve for } B \text{ and } C \text{ now } \\ A+B+C=1 \\ A+B-C=0 \\ \text{ so } A=\frac{-1}{4} \\ \frac{-1}{4}+B+C=1 \\ \frac{-1}{4}+B-C=0 \text{ adding equations together } \\ \frac{-2}{4}+2B=1 \\ 2B=\frac{3}{2} \\ B=\frac{3}{4} \\ \text{ then find } C \text{ using this system } \\ \text{ then you can finally find } D\]

  17. anonymous
    • one year ago
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    i got the integral! thank you!

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