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patbatE21

Evaluate the integral: Indefinite Integral ((x^3 + x^2 + 2x + 1) / (x^2 + 1)(x^2 + 2)) dx

  • 2 years ago
  • 2 years ago

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  1. patbatE21
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    |dw:1322597588202:dw|

    • 2 years ago
  2. patbatE21
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    so far I have x^3(A+C) + x^2 (B+D) + x(2A +C) +(2B + D)

    • 2 years ago
  3. patbatE21
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    But I believe I'm going in the wrong direction

    • 2 years ago
  4. slaaibak
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    \[{{x^3 + x^2 + 2x + 1}\over{(x^2 + 1)(x^2 + 2)} } = {{Ax + B}\over{x^2 + 1}} + {{Cx + D}\over{x^2 + 2}}\] By partial franctions. Now:

    • 2 years ago
  5. patbatE21
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    ok

    • 2 years ago
  6. patbatE21
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    I'm good so far

    • 2 years ago
  7. slaaibak
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    \[x^3 + x^2 + 2x + 1 = (Ax + B)(x^2 + 2) + (Cx + D)(x^2 + 1)\] \[x^3 + x^2 + 2x + 1 = Ax^3 + 2Ax + Bx^2 + 2B + Cx^3 + Cx + Dx^2 + D\] \[x^3 + x^2 + 2x + 1 = (A + C)x^3 + (B + D)x^2 + (2A + C)x + (2B + D)\] A + C = 1 B + D = 1 2A + C = 2 2B + D = 1

    • 2 years ago
  8. patbatE21
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    Alright...but the answer I saw was (1/2)ln (x^2 + 1) + (1 / square(2)) arctan(x/square(2)) + C

    • 2 years ago
  9. patbatE21
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    It's a frustrating problem

    • 2 years ago
  10. slaaibak
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    Lets take it further then: By solving A, B, C, D: A = 1 C=0 B=0 D=1

    • 2 years ago
  11. patbatE21
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    ok

    • 2 years ago
  12. slaaibak
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    \[{{x^3 + x^2 + 2x + 1}\over{(x^2 + 1)(x^2 + 2)}} = {x \over {x^2 + 1} } + {{1}\over{x^2 + 2}}\]

    • 2 years ago
  13. slaaibak
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    Now the fun starts! Now we integrate each one of those

    • 2 years ago
  14. patbatE21
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    ohhhhh ok

    • 2 years ago
  15. slaaibak
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    Now to integrate the first one: \[\int\limits {x \over {x^2 + 1} } dx\] \[x^2 = u\] \[2x dx = du\] \[xdx ={1\over2}du\] \[{1\over2}\int\limits{1\over{u + 1}} du\] =\[{1\over2}\ln(u + 1) + C\] resubstituting u with x^2: \[{1\over2}\ln(x^2 + 1) + C\]

    • 2 years ago
  16. slaaibak
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    Do you need help with the second part of the integral?

    • 2 years ago
  17. patbatE21
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    please if you don't mind

    • 2 years ago
  18. slaaibak
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    \[\int\limits{1\over{x^2 + 2}} dx\] now, by factoring out the 2: \[\int\limits {1\over{2({1\over2}x^2 + 1)}} dx\] \[{1\over2}\int\limits{1\over{({1\over2}x^2 + 1)}}dx\] Putting the 1/2 inside the bracket: \[{1\over2} \int\limits { {1}\over{ ({{x}\over{\sqrt2}})^2 + 1 } }dx\] \[{x \over {\sqrt 2} } = u\] \[dx = \sqrt2 du\] \[{1\over2}\int\limits {{\sqrt2}\over{u^2+1}} du\] \[{{\sqrt2}\over2}\int\limits {{du}\over{u^2+1}}\] = \[{\sqrt2 \over 2} \arctan u + c\] substituting back: \[{\sqrt2 \over 2} \arctan({x \over \sqrt2}) + C\] since \[{\sqrt2 \over 2 } = {1\over \sqrt2}\] \[{1\over \sqrt2} \arctan({x \over \sqrt2}) + C\]

    • 2 years ago
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