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Thread: int3egral , polinomials

  1. #1
    Boo
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    int3egral , polinomials

    Dear All!
    please, can someone help me solve this:

    $\displaystyle \int \frac{x^2+4}{x^2+16}dx$

    would it be proper:

    $\displaystyle }int \frac{x^2+16-12}{x^2+16}=\int \frac{x^2+16}{x^2+16dx}-\int \frac{12}{x^2+16}dx=\int(1)dx-12\int \frac{1}{x^2+16}$

    $\displaystyle =x-12 \frac{1}{4}arctan(x/4)+C$
    what to do now?
    Thanks in advance!
    Last edited by Boo; Jan 26th 2013 at 07:36 AM.
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  2. #2
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    Re: int3egral , polinomials

    It's OK to look up (or remember) a standard integral like this.

    If you want to do it yourself you can use the substitution $\displaystyle x=4\tan\theta$

    Haha. I think you finished the problem while I was typing my post.

    Apart from 12 x 1/4 = 3 of course, but really you're finished.
    Thanks from Boo
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  3. #3
    Boo
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    Re: int3egral , polinomials

    Thanks tutor!
    Still have to check if they ment sth else!!!
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  4. #4
    Boo
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    Re: int3egral , polinomials

    Hello!
    Please, the answer should be:
    $\displaystyle \frac{\sqrt{2}}{4}arctan\frac{x-\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{4}arctan\frac{ x+\sqrt{2}}{\sqrt{2}}+C$

    Please, the hell, how to get it???
    many thanks!!!
    Last edited by Boo; Jan 30th 2013 at 05:25 AM.
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  5. #5
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    Re: int3egral , polinomials

    Quote Originally Posted by Boo View Post
    Hello!
    Please, the answer should be:
    $\displaystyle \frac{\sqrt{2}}{4}arctan\frac{x-\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{4}arctan\frac{ x+\sqrt{2}}{\sqrt{2}}+C$

    Please, the hell, how to get it???
    many thanks!!!
    No. The answer you gave in your first post in this thread is correct.

    $\displaystyle x-3\tan^{-1}\left(\frac{x}{4}\right)+C$.
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  6. #6
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    Re: int3egral , polinomials

    The result you give is $\displaystyle \int \frac{x^2+4}{x^4+16} \,dx$ instead of $\displaystyle \int \frac{x^2+4}{x^2+16} \,dx$. Did you copy the question correctly in your original post?

    - Hollywood
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  7. #7
    Boo
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    Re: int3egral , polinomials

    Ujoj!!!
    The first one with $\displaystyle x^4$is right!!!
    Seems I could solve "my version" but I cant solve the one they worte in the book...
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  8. #8
    Boo
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    Re: int3egral , polinomials

    $\displaystyle \int \frac{x^2+4}{x^4+^16= }$
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  9. #9
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    Re: int3egral , polinomials

    You would factor $\displaystyle x^4+16$ as $\displaystyle (x^2+2\sqrt{2}x+4)(x^2-2\sqrt{2}x+4)$ and use partial fractions. It'll take a lot of work, but there's no trick or anything. Just follow the method and the answer should come right out.

    - Hollywood
    Thanks from Boo
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  10. #10
    Boo
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    Re: int3egral , polinomials

    $\displaystyle \frac{Ax+B}{x^2-2\sqrt{2}x+4}+\frac{Cx+D}{x^2-2\sqrt{2}x+4}$
    $\displaystyle (Ax+B)(x^2+2\sqrt{2}x+4)+(Cx+D)(x^2-2\sqrt{2}x+4)=x^2+4$
    $\displaystyle Ax^3+Bx^2+2\sqrt{2}Ax^2+2\sqrt{2}Bx+4Ax+4B+Bx^3+Cx ^2+2\sqrt{2}Bx^2+2\sqrt{2}Cx+4Bx+4C=x^2+4$

    Do we eliminate now x^3 and x^1 as it should be zero-I suppose?

    Am I right till now???
    Many thanks!!!
    Last edited by Boo; Jan 31st 2013 at 08:41 AM.
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  11. #11
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    Re: int3egral , polinomials

    That's the correct idea, but you're making mistakes. Here's what I got:

    $\displaystyle \frac{x^2+4}{x^4+16}=\frac{Ax+B}{x^2-2\sqrt{2}x+4}+\frac{Cx+D}{x^2+2\sqrt{2}x+4}$

    $\displaystyle (Ax+B)(x^2+2\sqrt{2}x+4)+(Cx+D)(x^2-2\sqrt{2}x+4)=x^2+4$

    $\displaystyle (A+C)x^3 + (2\sqrt{2}A - 2\sqrt{2}C + B + D)x^2 +$
    $\displaystyle (2\sqrt{2}B - 2\sqrt{2}D + 4A + 4C)x + (4B+4D) = x^2+4$

    We can equate all four coefficients:
    $\displaystyle A+C=0$
    $\displaystyle 2\sqrt{2}A - 2\sqrt{2}C + B + D=1$
    $\displaystyle 2\sqrt{2}B - 2\sqrt{2}D + 4A + 4C=0$
    $\displaystyle 4B+4D=4$

    Next you can solve for A, B, C, and D - four equations in four unknowns. You should get $\displaystyle A=C=0$ and $\displaystyle B=D=\frac{1}{2}$. Now you can integrate the two functions - do you know how?

    - Hollywood
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