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mrtuba

integral of (3x^3+2x^2+7x+13)/((x-1)^2(x^2+4)^2)

  • one year ago
  • one year ago

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  1. candrie
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    http://myalgebra.com/algebra_solver.aspx Plug your problem in here

    • one year ago
  2. mrtuba
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    \[\int\limits_{}^{}(3x^3+2x^2+7x+13)/((x-1)^2(x^2+4)^2)\]

    • one year ago
  3. myininaya
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    Did you try partial fractions Are is partial fractions what you need help on?

    • one year ago
  4. myininaya
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    oops or* lol sorry

    • one year ago
  5. mrtuba
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    i did partial fractions but its so incredibly terrible that the expanded for after cross multiplication took up to line of college rule paper written in small print. I feel like there is some simplifying i could do before doing partial fractions

    • one year ago
  6. myininaya
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    It does look like it can nasty but I don't see any other way to go about it but... just because i don't see any other way doesn't mean it doesn't exist

    • one year ago
  7. myininaya
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    \[\frac{3x^3+2x^2+7x+13}{(x-1)^2(x^2+4)^2}\] \[=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+4}+\frac{Ex+F}{(x^2+4)^2}\] Yeah this is going to be ugly

    • one year ago
  8. myininaya
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    So you didn't find those constant values for A,B,C,D,E and F right?

    • one year ago
  9. mrtuba
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    no i just saw the set up after cross multiplying and expanding looked so crazy that i thought i had to have missed something that could make it easier.

    • one year ago
  10. myininaya
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    \[=\frac{A(x-1)(x^2+4)^2}{(x-1)^2(x^2+4)^2}+\frac{B(x^2+4)^2}{(x-1)^2(x^2+4)^2}+\frac{(Cx+D)(x-1)^2(x^2+4)}{(x-1)^2(x^2+4)^2}+\frac{(Ex+F)(x-1)^2}{(x^2+4)^2(x-1)^2}\] \[3x^3+2x^2+7x+13=A(x-1)(x^2+4)^2+B(x^2+4)^2+ \] \[(Cx+D)(x-1)^2(x^2+4)+(Ex+F)(x-1)^2 \] And then continue from here ....

    • one year ago
  11. mrtuba
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    yeah i went past that, just looks like this one will be ugly i guess ill just keep chugging thanks.

    • one year ago
  12. myininaya
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    Sorry... I'm not able to offer an easier way :(

    • one year ago
  13. myininaya
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    I don't think there is one...

    • one year ago
  14. malevolence19
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    Plug in X=1 to get a value, x=0 might also help

    • one year ago
  15. mrtuba
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    can not figure out the system of equations this leads me to

    • one year ago
  16. agentx5
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    1. Carefully check your + and - signs in the original problem. You've currently an unfactorable polynomial in the numerator with a larger, higher degree polynomial in the denominator. I'd agree with @myininaya's approach. One missed sign could make it factorisable such that something could simplify/cancel out and the whole thing become much more feasible. 2. There isn't an easier way, that I can see either, you're just going to have to solve that system of equations, namely you're going to have to use both methods of combining systems of equations and making substitutions, AND using known vertical asymptote values for the denominator (in this case +1 and 2i). Yes, I'm serious, use an imaginary root here, you kind of have to. Post your system of equation here, and try those roots, see what happens (things cancel out). 3. Using that I got: A = 0 B = 1 C = 0 D = -1 E = 1 F = 1 You'll also need to make use of this fact: \[\frac{d}{dx}(\frac{1}{a} \tan^{-1}(\frac{x}{a})) = \frac{1}{x^2+a^2}\] This problem as written is do-able, but it's a lot of work and will also most likely also force you to remember some anti-derivatives which have trig function forms and make use of half-angle and double-angle formulas too. So you got a lot of work to do, better get to it eh? ;-D

    • one year ago
  17. eliassaab
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    Here is a summary and a conclusion to what have been above and what have not been \[ \frac{3 x^3+2 x^2+7 x+13}{(x-1)^2 \left(x^2+4\right)^2}=\frac {x+1}{\left(x^2+4\right)^2} -\frac{1}{x^2+4}+\frac{1}{( x-1)^2}\\ \int \left(\frac{x+1}{\left(x^2+4\right)^2}-\frac{1}{x^2+4}+\frac{1}{(x-1)^2}\right) \, dx=\\\frac{x-4}{8 \left(x^2+4\right)}-\frac{1}{x-1}-\frac{7 }{16} \tan ^{-1}\left(\frac{x}{2}\right)+ C \]

    • one year ago
  18. agentx5
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    @eliassaab is 100% correct :-)

    • one year ago
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