anonymous
  • anonymous
Please help! :) Write the partial fraction decomposition of (3x^2 + 4)/(x^2 + 1)^2
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\frac{ 3x^2 + 4 }{ (x^2 + 1)^2 } = \frac{ Ax + B }{ x^2 +1 } + \frac{ Cx + D }{ (x^2 + 1)^2 }\]
anonymous
  • anonymous
You need to add them (which will require identical denominators, obtained by manipulation) and solve for A, B, C and D.
anonymous
  • anonymous
I'm not sure what to do.

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anonymous
  • anonymous
To make denominators same: \[\frac{ Ax + B }{ x^2 + 1 } \times \frac{ x^2 + 1 }{ x^2 + 1 } + \frac{ Cx + D }{ (x^2+1)^2 }\] \[= \frac{ Ax^3 + Ax + Bx^2 + B + Cx + D }{ (x^2+1)^2 } = \frac{ 3x^2 + 4 }{ (x^2 + 1)^2 }\] This implies that: Ax^3 + Ax + Bx^2 + B + Cx + D = 3x^2 + 4 By grouping all like terms of the polynomial to compare with the original numerator: Ax^3 = 0x^3 ---> A = 0 Bx^2 = 3x^2 ---> B = 3 Ax + Cx = 0x --> A = -C = 0 B + D = 4 ------> D = 1 so plugging this back in: \[\frac{ Ax + B }{ x^2 + 1 } + \frac{ Cx + D }{ (x^2 + 1)^2 } = \frac{ 3 }{ x^2 + 1 } + \frac{ 1 }{ (x^2 + 1)^2 }\] This means that: \[\frac{ 3 }{ x^2 + 1 } + \frac{ 1 }{ (x^2 + 1)^2 }\] is equivalent to \[\frac{ 3x^2 + 4 }{ (x^2 + 1)^2 }\]

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