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Math Help - integral

  1. #1
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    integral

    Hi everyone,

    Could someone please explain to me how to solve the integral x^3-x^2-x+1/(x^2+9)

    I know that I have to do long divition first.

    After doing this I got x-1-10x+10/x^2+9, but the correct answer for this step is x-1-8x-1/x^2+9. Does anyone see where I went wrong?

    After that I am supposed to use a table and the correct answer is

    1/2x^2-x-8(1/2ln(x^2+9))+8(1/3tan^-1(x/3))+c

    Can someone explain to me how they got this as their final answer, please?

    Thank you very much
    Last edited by chocolatelover; September 16th 2007 at 02:41 PM.
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  2. #2
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    Quote Originally Posted by chocolatelover View Post
    x^3+x^2+x+1/x^2+9
    First, please fix your notation. I think you mean: (x^3+x^2+x+1)/(x^2+9). Remember studying the Order of Operations? It makes a difference.
    After doing this I got x - 1 - (10x+10)/(x^2+9)
    Nope. Both those '10's should be '8's. Since you didn't show your work, it's a little tough to spot your error. Just try it again and be mroe careful.

    It may help for you to notice that the numerator can be factored, but probably not. You should keep your eyes open for interesting things.
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  3. #3
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    Hi,

    The original problem is: x^3-x^2-x+1/(x^2+9)

    I then did long division and got x-1-(10x+10)/(x^2+9)

    I said that x^2+9 goes into x^3-x^2 X times. Then, after subtracting the first row from the top row I got -x^2-10x+1 and said that that went into the original equation -1 times. I then got -10x+10 as a remainder. Do you see the error?

    Thank you
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  4. #4
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    Still needs parentheses in the numerator to mean what you want.

    You are going to have to make up your mind. Are those signs in the numerator positive or negative? If they are negative, you may be on the right track. If they are poositive, as you posted originally, then that is where you wandered off.
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  5. #5
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    This is the original problem:

    integral (x^3-x^2-x+1)/(x^2+9)
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  6. #6
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    Okay. Then you shold get 10x - 8, not 10x - 10. There's a sign problem in there, somehwere.
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  7. #7
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    Long division is not well done.

    We have \int\frac{x^3-x^2-x-1}{x^2+9}\,dx=\int\left(x-1+\frac{8-10x}{x^2+9}\right)\,dx

    Play a bit with this.
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  8. #8
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    I've checked and re-checked it and I still can't find the mistake.

    I'm supposed to get the integral. x-1-8x-1/(x^2+9)

    I still get -10x+10. I have -10x-0x, which is -10x and 1--9 which is 10.

    Can someone help me with this, please? I also don't know what to do after I do the long division correctly. I'm supposed to use the table integral du/a^2+u^2=1/atan^-1u+c

    I understand that from the integral x-1-8x-1/x^2+9, you would get 1/2x^2-x-(8)(1/2lnx^2+9), but I don't understand how you know to add 8 and do the following+ 8(1/3 tan^-1(x/3))+c

    Thanks
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  9. #9
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    Can someone help me on this, please?
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    I agree with chocolatelover, the remainder from the long division is 10 - 10x:

    \int \frac {x^3 - x^2 - x + 1}{x^2 + 9}~dx = \int \left( x - 1 + \frac {10 - 10x}{x^2 + 9} \right)~dx

    ................................ = \int \left(x - 1 + \frac {10}{x^2 + 9} - \frac {10x}{x^2 + 9} \right)~dx

    for the third term, use the formula: \int \frac {du}{a^2 + u^2} = \frac 1a \arctan \left( \frac ua \right) + C

    for the last term, use the substitution (not integration by parts, regular substitution) u = x^2 + 9

    can you take it from here, chocolatelover?
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  11. #11
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    Yes, now I see what I need to do. Thank you very much

    except I got x-1(-10x-1)/(x^2+9)

    How did you get a positive 10x?
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Yes, now I see what I need to do. Thank you very much

    except I got x-1(-10x-1)/(x^2+9)

    How did you get a positive 10x?
    i didn't get a positive 10x. i got 10 - 10x, which is what you got at first
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  13. #13
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    chocolatelover, you changed the question several times!

    Well, anyway doesn't affect to the problem.

    Remember that \int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+k, it's useful when you have integrals like \int\frac x{x^2+a}\,dx
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