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Math Help - Integrals

  1. #1
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    Integrals

    I know the basics of integrals but I am having problems with division with these two problems:
    1. indefinite... (x^2+x-2)/(x^2) The answer is x+ln |x| + 2x^-1+C
    =x^2+x-2+x^-2
    =((x^3/3)+(x^2/2)-(2x)+(x^-1/-1)
    =((1/3)(x^3)+(1/2)(x^2)-(2x)-(x^-1) at least my solution. I know I mess this one up.
    2. indefinite...((1-x^2)^1/2)-1/(1-x^2)^1/2) The answer is x-(sin(x))^-1+C. Also, I am aware that 1/(1-x^2)^1/2 is (sin x)^-1
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  2. #2
    Grand Panjandrum
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    [quote=driver327;64097]I know the basics of integrals but I am having problems with division with these two problems:
    1. indefinite... (x^2+x-2)/(x^2) The answer is x+ln |x| + 2x^-1+C
    =x^2+x-2+x^-2
    =((x^3/3)+(x^2/2)-(2x)+(x^-1/-1)
    =((1/3)(x^3)+(1/2)(x^2)-(2x)-(x^-1) at least my solution. I know I mess this one up.

    \int \frac{x^2+x-2}{x^2}~dx=\int 1+1/x-2/x^2~dx = x+\ln(x)+2/x + C

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by driver327 View Post
    I2. indefinite...((1-x^2)^1/2)-1/(1-x^2)^1/2) The answer is x-(sin(x))^-1+C. Also, I am aware that 1/(1-x^2)^1/2 is (sin x)^-1
    This is ambiguous, but I will assume you mean:

    \int \frac{(1-x^2)^{1/2}-1}{(1+x^2)^{1/2}}~dx=\int 1-\frac{1}{(1+x^2)^{1/2}}~dx

    and the rest you should be able to do.

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    This is ambiguous, but I will assume you mean:

    \int \frac{(1-x^2)^{1/2}-1}{(1+x^2)^{1/2}}~dx=\int 1-\frac{1}{(1+x^2)^{1/2}}~dx

    and the rest you should be able to do.

    RonL
    On the second part, I see now how it becomes x-sin(x)^-1, but I am a bit confused on how it goes from point A to point B(in the quotes). I know this may be stupid, but i really do not know.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by driver327 View Post
    On the second part, I see now how it becomes x-sin(x)^-1, but I am a bit confused on how it goes from point A to point B(in the quotes). I know this may be stupid, but i really do not know.
    It's division. Think of it this way:

    Let b = (1 - x^2)^{1/2}. Then you have:

    \frac{b - 1}{b} = \frac{b}{b} - \frac{1}{b} = 1 - \frac{1}{b} = 1 - \frac{1}{(1 - x^2)^{1/2}}

    -Dan
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  6. #6
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    I must have overlooked that, thanx
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