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Kelumptus

  • 2 years ago

Is the following a correct solution?

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  1. Kelumptus
    • 2 years ago
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    \[\int\limits_{}^{}\frac{ 1 }{ 4+x ^{2} }dx=\frac{ (x-\sqrt{4}i) }{ (x+\sqrt{4}i) }\] \[=\int\limits_{}^{}\frac{u-2\sqrt{4}i}{u}=\int\limits_{}^{}\frac{-2\sqrt{4}i}{u}=-2\sqrt{4}i \ln|u| + c\] \[=-2\sqrt{4}i \ln|x+\sqrt{4}i| + c\]

  2. shubhamsrg
    • 2 years ago
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    i didnt get your solution/or the question..please clarify..

  3. Kelumptus
    • 2 years ago
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    The question is: is \[-2\sqrt{4}i \ln |x+\sqrt{4}i| + c\] the indefinite integral of \[\int\limits_{}^{}\frac{1}{4+x ^{2}} dx\]

  4. shubhamsrg
    • 2 years ago
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    you can see 1/ 4+x^2 = 1/ (x+ 2i)(x-2i) = 1/4i * (x+2i - (x-2i))/(x+2i)(x-2i) now separate(simplify) and integrate..hope that helped.. am sorry i dint get your soln much.

  5. mahmit2012
    • 2 years ago
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    |dw:1354951402025:dw|

  6. Kelumptus
    • 2 years ago
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    That is correct also, but i am wondering if the solution involving complex numbers is also correct.

  7. Kelumptus
    • 2 years ago
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    Actually, the solution that you typed should be 1/2 Tan(x/2) + K

  8. Kelumptus
    • 2 years ago
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    Ahh crap, sorry, i didn't read your solution properly, i just assumed that you were using trig substitution. Forget what i just said, let me take a closer look =)

  9. mahmit2012
    • 2 years ago
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    yes, I can show that both functions has a same power series, and it shows the equality is corect.

  10. mahmit2012
    • 2 years ago
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    for example tan|dw:1354952178836:dw|

  11. mahmit2012
    • 2 years ago
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    for example tan^-1(i)=inf

  12. Kelumptus
    • 2 years ago
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    Ahh, thats using the power series. Cheers =)

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