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Math Help - integral ...

  1. #1
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    Smile integral ...

    What is the problem here I get diffrent result ....

    \int_0^2 \frac{\frac{5}{2}}{4x^4+25}= \frac{5}{2}\int_0^2 \frac{1}{(2x^2)^2+5^2}dx = \frac{5*1}{2*5} arctan (\frac{2x^2}{5})
    Last edited by gilyos; January 11th 2010 at 06:20 AM.
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  2. #2
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    Quote Originally Posted by gilyos View Post
    What is the problem here I get diffrent result ....

    \int_0^2 \frac{\frac{5}{2}}{4x^4+25}= \frac{5}{2}\int_0^2 {1}{(2x^2)^2+5^2}dx = \frac{5*1}{2*5} arctan (\frac{2x^2}{5}
    I think you mean \frac{1}{(2x^2)^2+5^2} not {1}{(2x^2)^2+5^2}

    Look to this formula carefaully (Assuming a dont equal 0):
    \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} arctan \frac{x}{a} + C

    Are you sure your problem is similar to it?
    i.e. when you substituting u=2x^2 , You will get your du in the numerator?

    You forgot the dx in the first integral. Also its a definite integral, Its result must be a number not a function.
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  3. #3
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    I check and the result before the answer and afther the answer is different ... why ?
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  4. #4
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    I told you : you have a definite integral, its result must be a number not a function.

    Did you substitute u=2x^2 ?
    Try it and you will see what I mean.
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  5. #5
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    You can use partial fractions:
    4x^4+25 = (2x^2)^2+20x^2+5^2-20x^2 = (2x^2+5)^2-(\sqrt{20}x)^2 =
    = (2x^2-\sqrt{20}x+5)(2x^2+\sqrt{20}x+5)
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by gilyos View Post
    What is the problem here I get diffrent result ....

    \int_0^2 \frac{\frac{5}{2}}{4x^4+25}= \frac{5}{2}\int_0^2 \frac{1}{(2x^2)^2+5^2}dx = \frac{5*1}{2*5} arctan (\frac{2x^2}{5})
    Quote Originally Posted by Unbeatable0 View Post
    You can use partial fractions:
    4x^4+25 = (2x^2)^2+20x^2+5^2-20x^2 = (2x^2+5)^2-(\sqrt{20}x)^2 =
    = (2x^2-\sqrt{20}x+5)(2x^2+\sqrt{20}x+5)
    Strange. This seems to be exactly what I suggested to you last time you posted this question.
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  7. #7
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    Quote Originally Posted by Drexel28 View Post
    Strange. This seems to be exactly what I suggested to you last time you posted this question.
    Oh sorry I didn't notice. I found it now here
    http://www.mathhelpforum.com/math-he...rp-8-11-a.html
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Unbeatable0 View Post
    Oh sorry I didn't notice. I found it now here
    http://www.mathhelpforum.com/math-he...rp-8-11-a.html
    Not at all your fault of course!
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