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Math Help - Easy integral!!

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    Easy integral!!

    Solve the following integral
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    Quote Originally Posted by matty888 View Post
    Solve the following integral
    Use a partial fraction decomposition.
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  3. #3
    Moo
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    Strange thing because it's not defined in 1...
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    Quote Originally Posted by Moo View Post
    Strange thing because it's not defined in 1...
    Good spot. You get +oo when treating it as an improper integral and taking the limit .....
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    Quote Originally Posted by mr fantastic View Post
    Good spot. You get +oo when treating it as an improper integral and taking the limit .....
    Yep... The integral doesn't exist because the integrand isn't continuous ~
    I wonder if it's a typo or whatever..
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    Thanks guys,I was asked if it converges and wasnt sure about what i was getting(0),so from ur reckoning then it will diverge,would I be right in saying that??
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    Quote Originally Posted by matty888 View Post
    Thanks guys,I was asked if it converges and wasnt sure about what i was getting(0),so from ur reckoning then it will diverge,would I be right in saying that??
    Yes.
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    Quote Originally Posted by matty888 View Post
    Solve the following integral
    \int_0^{1}\frac{dx}{1-x^2}=arctanh(x)\bigg|_0^{1}\to\infty[/tex]


    Alternate integration method
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by matty888 View Post
    Solve the following integral
    Also \forall{x}\in(0,1],\frac{1}{1-x^2}>\frac{1}{x^2}

    So then \int_0^{1}\frac{1}{x^2}dx\leq\int_0^{1}\frac{1}{1-x^2}dx

    \int_0^{1}\frac{dx}{x^2} is a divergent special case p-series test

    So from that we see that \int_0^{1}\frac{dx}{1-x^2} diverges
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    Moo
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    Quote Originally Posted by Mathstud28 View Post
    Also \forall{x}\in(0,1],\frac{1}{1-x^2}>\frac{1}{x^2}

    [spip]
    So it implies that : 1-x^2<x^2 \Longleftrightarrow 2x^2>1, for x \in [0,1].

    Now, tell me, am I misunderstanding you or are you just looking for something that would help you ?
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    So it implies that : 1-x^2<x^2 \Longleftrightarrow 2x^2>1, for x \in [0,1].

    Now, tell me, am I misunderstanding you or are you just looking for something that would help you ?
    I am not going to deny it this time. My calculator lied to me and I did not check it. My TI-89 has been acting up lately...sorry for the confusion
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