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Blisk

integral(1/((2x)^2+1)dx

  • 2 years ago
  • 2 years ago

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  1. Blisk
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    show full steps if possible o_o

    • 2 years ago
  2. uber1337h4xx0r
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    Ready to work together on this?

    • 2 years ago
  3. uber1337h4xx0r
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    This one's going to be a bit tricky, I'm afraid.

    • 2 years ago
  4. Blisk
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    lol i know its pretty rediculous

    • 2 years ago
  5. uber1337h4xx0r
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    Okie dokie. Let's do this. I'll give you a hint: The thing is in the form of a trig identity.

    • 2 years ago
  6. uber1337h4xx0r
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    Can you tell me the trig identities for cos^-1 tan^-1 and sin^-1 first? After that, I think you'll know exactly what the answer is :D

    • 2 years ago
  7. galacticwavesXX
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    don't you have to find the antiderivative of the function?

    • 2 years ago
  8. Blisk
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    well i know its arctan its just the steps involved inbetween to find the answer

    • 2 years ago
  9. Blisk
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    im just not sure if its u sub or something thats why i wana see it with full steps

    • 2 years ago
  10. Blisk
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    @galactic yes you have to find the antiderivative saying integral is another way of saying anti derivative

    • 2 years ago
  11. uber1337h4xx0r
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    You know your "X" is 2x. Because of that, you must chain rule it. When you do the derivative of 2x, you get 2. Pull that two out, and because it's on the bottom, it turns to 1/2. So 1/2 tan^-1 (2x) + c should be the answer. But double check with other people to be safe.

    • 2 years ago
  12. uber1337h4xx0r
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    If my wording was weird, I meant that normally it'd be just tan^-1 if x^2 was there. But we have (2x)^2 instead of just (x)^2.

    • 2 years ago
  13. Blisk
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    shouldnt the 1/2 go on the top and turn into 2arctan(2x)?

    • 2 years ago
  14. galacticwavesXX
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    @uber i wanted to ask how comes we are using the trig identities. is it because it is one over the function??

    • 2 years ago
  15. uber1337h4xx0r
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    @Blisk: Well I mean it's originally just (1/(2(tan^-1(2x)). So you pull the 1/2 to the left.

    • 2 years ago
  16. uber1337h4xx0r
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    @galactic; Look at the derivative of tan^-1.

    • 2 years ago
  17. Blisk
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    how did you get 1/(2tan^-1(2x))?

    • 2 years ago
  18. daveboyback
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    Can Someone Help Me After They finish doing this guy!! :D

    • 2 years ago
  19. Blisk
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    that sounds wrong lol

    • 2 years ago
  20. daveboyback
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    ok i know :D

    • 2 years ago
  21. uber1337h4xx0r
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    @blisk: I chain ruled it.

    • 2 years ago
  22. daveboyback
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    Sory

    • 2 years ago
  23. uber1337h4xx0r
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    d/dx (2x) = 1

    • 2 years ago
  24. Blisk
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    i didnt know you could chain rule backwards

    • 2 years ago
  25. uber1337h4xx0r
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    Oh shoot! You're right! I think I made up a rule then. :(

    • 2 years ago
  26. galacticwavesXX
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    isnt the derivative of tan^-1......sec^2(x)tan^-1?!?

    • 2 years ago
  27. Blisk
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    the deriv of tan^-1 is 1/x^2+1

    • 2 years ago
  28. Blisk
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    the deriv of tan = sec^2(x) and the deriv of sec is sectan

    • 2 years ago
  29. galacticwavesXX
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    your right i confused myself sorry!!!

    • 2 years ago
  30. Blisk
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    lol dont worry about it

    • 2 years ago
  31. abtrehearn
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    Draw a right triangle having legs of lengths 1 and 2x. Place the unit length segment om the positive x-axis with one end at the origin O. Erect the perpendicular to the x-axis at A(1, 0) and place the other leg on that perpendicular with its lower end at (1, 0). The other end of that vertical leg is at B(1, 2x). Lrt t be angle AOB. Then tan(t) = 2x/1 = 2x, and (sec(t))^2 dt = 2 dx =>dx = (sec(t))^2/2 dt. If we substitute into the integral, we get(1/2)(sec(t))^2 dt/(1 + (tan(t))^2. and we end up integrating dt/2, so that simplifies to t/2. Getting back to original variable x, we have (1/2) Arctan(2x) + C for the final answer.

    • 2 years ago
  32. Blisk
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    would this work too? S = integral S1/((2x)^2+1) dx u = 2x du = 2 dx => 1/2 du = dx S(1/u^2+1)(1/2) du 1/2S(1/u^2+1) du 1/2 tan^-1(u) Plug U back in 1/2 tan^-1(2x) is that also correct abtre?

    • 2 years ago
  33. abtrehearn
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    Yes it is.

    • 2 years ago
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