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amorfide

I need to know how to integrate this

  • one year ago
  • one year ago

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  1. amorfide
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    • one year ago
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  2. mukushla
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    use partial fractions here\[\frac{3x+4}{(x^2+4)(x-3)}=\frac{Ax+B}{x^2+4}+\frac{C}{x-3}\]and find A,B and C

    • one year ago
  3. carson889
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    Split using partial fractions: \[\frac{ 3x+4 }{ (x ^{2}+4)(x-3) } = \frac{ A }{ x-3 } + \frac{ Bx + C}{ (x ^{2} + 4) } = \frac{ A(x^{2} +4)+(Bx+C)(x-3) }{ (x ^{2}+4)(x-3) }\] Set equal to 3, looking at just numerator: \[3*3 + 4 = 13 = A(9+4) + (3B+C)(0) = 13A \rightarrow A = 1\] Set x equal to 0: \[0 + 4 = 4 = 4A + C \rightarrow 4 = 4+C \rightarrow C = 0\] Set x equal to 1: \[3 + 4 =A(1+4) + (B+C)(1-3) = 5A -2B -2C \rightarrow 5 - 2B - 0 \rightarrow 2 = -2B \rightarrow B = -1\] Therefore: \[\frac{ 3x+4 }{ (x ^{2}+4)(x-3)} = \frac{ 1 }{ x-3 } - \frac{ x }{ x ^{2} + 4 }\] Now integrate those.

    • one year ago
  4. amorfide
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    thank you very much, you have no idea how much i love you xD

    • one year ago
  5. carson889
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    No problemo The integral of 1/(x-3) is ln(x-3) and the integral of -x/(x^2 + 4): u-sub: u = x^2 + 4, du = 2x dx \[\frac{ -1 }{ 2 } \int\limits_{}^{}\frac{ 1 }{ u }du = \frac{ -1 }{ 2 }\ln(u) = \frac{ -1 }{ 2 }\ln(x ^{2}+4)\] \[\left( \ln(2-3) - \frac{ 1 }{ 2 }\ln(4+4) \right)-\left( \ln(0-3) - \frac{ 1 }{ 2 }\ln(0+4) \right) \] \[= \ln(-1)-\frac{ 1 }{ 2 }\ln(8)-\ln(-3)-\frac{ 1 }{ 2 }\ln(4) = \ln(\frac{ 1 }{ 3\sqrt{2} })\]

    • one year ago
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