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  1. ksaimouli
    • one year ago
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    @mathmate

  2. ksaimouli
    • one year ago
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    i did this with u substituion it did not work

  3. ksaimouli
    • one year ago
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    i am getting something like ln2/2

  4. mathmate
    • one year ago
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    It looks to me more an atan(2x)

  5. ksaimouli
    • one year ago
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    how

  6. ksaimouli
    • one year ago
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    i took u=1+4t^2

  7. ksaimouli
    • one year ago
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    du/8t=dt

  8. mathmate
    • one year ago
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    The standard integral is \( \int \frac{dx}{1+x^2} = atan(x) + C \)

  9. ksaimouli
    • one year ago
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    |dw:1358025284718:dw|

  10. mathmate
    • one year ago
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    So you'd be integrating \( \int \frac{du}{8u(1+4t^2)} \)

  11. ksaimouli
    • one year ago
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    use the equation button

  12. mathmate
    • one year ago
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    The result is not ln(u)

  13. ksaimouli
    • one year ago
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    ok i should remember that formula right?

  14. mathmate
    • one year ago
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    I suggest you use y=2t

  15. mathmate
    • one year ago
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    Yes, it's on the first page of standard integrals. Check it out.

  16. ksaimouli
    • one year ago
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    ok

  17. mathmate
    • one year ago
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    \( \large \int_0^{1/2} \frac{dt}{1+4t^2}= \int_0^{1} \frac{2dy}{1+y^2} = [atan(y)]_0^{1}= 2\pi/4=\pi/2 \)

  18. ksaimouli
    • one year ago
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    @mathmate i found the formula to be \[\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a }+c\]

  19. mathmate
    • one year ago
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    Right, so a=1/2.

  20. ksaimouli
    • one year ago
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    what is x then

  21. ksaimouli
    • one year ago
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    \[\int\limits_{}^{}\frac{ 1 }{ a^2+x^2}=\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a}+c\]

  22. mathmate
    • one year ago
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    x is your t. you can replace the identity with \( \large \int \frac{dx}{a^2+x^2} = \int \frac{dt}{a^2(1+(t/a)^2)} \)

  23. mathmate
    • one year ago
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    and a=1/2

  24. ksaimouli
    • one year ago
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    can i do \[\frac{ 1 }{ 1 }\tan^{-1} \frac{ 2t }{ 1}\]

  25. ksaimouli
    • one year ago
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    and t=1/2

  26. mathmate
    • one year ago
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    t is a variable, and you cannot equate it to a constant. Use the formula you found, change t for x, and 1/2 for a, then the left-hand side is exactly your question. All you need to do after that is to evaluate the right-hand side.

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