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KROD53

  • 3 years ago

How do you integrate (x-1)/(x^2 -4x + 5)?

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  1. bilgads
    • 3 years ago
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    \[\int\limits_{ }^{} \frac{ x-1 }{x ^{2}-4x+5 } \]

  2. bilgads
    • 3 years ago
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    \[\frac{ 1 }{2 } \int\limits_{}^{} \frac{ 2x-4+2 }{ x^{2}-4x+5 } \]

  3. bilgads
    • 3 years ago
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    \[\frac{ 1 }{ 2 } \int\limits_{}^{} \left[ \int\limits_{}^{} \frac{ 2x-4 }{ x ^{2}-4x+5 } dx + \int\limits_{}^{} \frac{ 2 }{ x ^{2}-4x+5 }dx\right] \]

  4. bilgads
    • 3 years ago
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    sorry we have not integral side \[\frac{ 1 }{ 2 }\]

  5. bilgads
    • 3 years ago
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    \[\frac{ 1 }{ 2 } \left[ \ln (x ^{2}-4x+5) + 2\int\limits_{}^{}\frac{ 1 }{ x ^{2} -4x+5 } dx\right]\]

  6. bilgads
    • 3 years ago
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    we have \[\int\limits_{}^{} \frac{ 1 }{ (x-2)^{2} +1}dx = \tan^{-1} (x-2)\]

  7. bilgads
    • 3 years ago
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    so the integral is \[\frac{ 1 }{ 2 }\ln (x ^{2}-4x+5) + \tan^{-1} (x-2) +C\]

  8. adi171
    • 3 years ago
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    \[\int\limits x-1\div x ^{2}-4x + 4 +4 -8 +5\]

  9. adi171
    • 3 years ago
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    \[\int\limits \frac{ x-1 }{ (x-2)^2 +1 }\]

  10. adi171
    • 3 years ago
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    \[\int\limits \frac{ x }{ (x-2)^2 +1 }dx - \int\limits \frac{ 1 }{ (x-2)^2 +1 } dx\]

  11. adi171
    • 3 years ago
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    \[\ln (x^2 -4x +5) - \tan^{-1} (x-2) +C\]

  12. adi171
    • 3 years ago
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    this is the right ans....when u have a quadratic eq. in denominator remember to add and subtract b^2/2a.....and then make quadratic + constant form eq.

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