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anonymous

  • one year ago

How to integrate the following: 2x/(x^3-1)

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  1. anonymous
    • one year ago
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    \[I=\int\limits \frac{2 x }{\left( x-1 \right)\left( x^2+x+1 \right) }dx\] \[\frac{ 2x }{\left( x-1 \right)\left( x^2+x+1 \right) }=\frac{ A }{ x-1 }+\frac{ Bx+C }{ x^2+x+1 }\] \[2x=A \left( x^2+x+1 \right)+\left( Bx+C \right)\left( x-1 \right)\] equate terms like powers of x 0=A+B 2=A-B+C 0=A-C find A,B,C and integrate.

  2. anonymous
    • one year ago
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    can you complete it?

  3. anonymous
    • one year ago
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    http://prntscr.com/8an7mo this term is much complicated

  4. anonymous
    • one year ago
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    Thanks Surjithayer. Your reply was very helpful and I could complete it.

  5. anonymous
    • one year ago
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    yw

  6. anonymous
    • one year ago
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    Unable to get correct Answer. Can you further solve To final answer.

  7. anonymous
    • one year ago
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    A+B=0,B=-A A-C=0 C=A 2=A+A+A \[A=\frac{ 2 }{ 3 },B=-A=-\frac{ 2 }{ 3 },C=A=\frac{ 2 }{ 3 }\]

  8. anonymous
    • one year ago
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    \[I=\int\limits \frac{ \frac{ 2 }{ 3 } }{ x-1 }dx+\int\limits \frac{ \frac{ -2 }{ 3 }x+\frac{ 2 }{ 3 } }{ x^2+x+1 }dx\] \[=\frac{ 2 }{ 3 }\ln \left( x-1 \right)-\frac{ 1 }{ 3 }\int\limits \frac{ 2x-2 }{ x^2+x+1 }dx\] \[=\frac{ 2 }{ 3 }\ln \left( x-1 \right)-\frac{ 1 }{ 3 }\int\limits \frac{ 2x+1-3 }{ x^2+x+1 }dx\] \[=\frac{ 2 }{ 3 }\ln \left( x-1 \right)-\frac{ 1 }{ 3 }\int\limits \frac{ 2x+1 }{ x^2+x+1 }dx+\int\limits \frac{ 1 }{x^2+x+1 }dx\]

  9. anonymous
    • one year ago
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    Thanks Surjithayer now it is clear.

  10. anonymous
    • one year ago
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    yw

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