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Math Help - Integrals!

  1. #1
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    Integrals!

    Hi people!

    I can't solve these ones, they're tough!

    Solve \int {\frac{{\sqrt {\sqrt {x^4 + 1} - x^2 } }}<br />
{{x^4 + 1}}\,dx}

    Evaluate \int_0^1 {x\ln x\,dx}

    Thanks!
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  2. #2
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    for the second one do integration by parts, you should get an answer of

    \frac{x^2 lnx}{2} - \frac{x^2}{4}
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  3. #3
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    The definition of "tough" always has eluded me.

    Frequent Definition: I don't know how to do it.

    Occasional Definition: I can do it, but it takes too long and requires too much effort.

    Pratical Definition: It can be done, but who would want to?

    Reasonable Definition: I could spend a lot of time on this, using some way that I have concocted, but even if it works, there simply MUST be an easier way.

    This leads me to ask two things:

    1) Why have you been given these integrals? Are you REALLY expected to solve them in a closed form? Without a very specific hint, and with your missing he Integration by Parts on the second, it seems to me that the first is not reasonable. Do you have a specific hint?

    2) Does one get to use numerical methods? Numbers are not evil. It's okay if that's what we have.
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  4. #4
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    Quote Originally Posted by Mess View Post
    Hi people!

    I can't solve these ones, they're tough!

    Solve \int {\frac{{\sqrt {\sqrt {x^4 + 1} - x^2 } }}<br />
{{x^4 + 1}}\,dx}

    Evaluate \int_0^1 {x\ln x\,dx}

    Thanks!
    For the first one, try substituting x^2 = tan(t).

    You will end up with an integrand looking like:

    \frac{\sqrt{sec(t)-tan(t)}}{sec^2(t)}
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  5. #5
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    Here's a specific calculation in this case.

    Quote Originally Posted by Mess View Post

    Evaluate \int_0^1 {x\ln x\,dx}
    \ln x=\int_1^x\frac1u\,du. So,

    \int_0^1 {x\ln x\,dx} = - \int_0^1 {\int_x^1 {\frac{x}<br />
{u}\,du} \,dx} = - \int_0^1 {\underbrace {\int_0^u {x\,dx} }_{\frac{1}<br />
{2}u^2 }\frac{1}<br />
{u}\,du} .

    The rest is routine, and the answer is -\frac14.

    Even I'd like to take a more general case:

    Let m,n>0.

    Define \int_0^1x^m\ln^nx\,dx.

    Make substitution u=-\ln x which can be written as x=e^{-u}\implies dx=-e^{-u}\,du, the integral becomes

    \int_0^1 {x^m \ln ^n x\,dx} = ( - 1)^n \int_0^\infty {e^{ - (m + 1)u} u^n \,du} .

    Define another substitution according to

    \alpha = (m + 1)u \implies u = \frac{\alpha }<br />
{{m + 1}}\,\therefore \,du = \frac{1}<br />
{{m + 1}}\,d\alpha ,

    \int_0^1 {x^m \ln ^n x\,dx} = \frac{{( - 1)^n }}<br />
{{(m + 1)^{n + 1} }}\int_0^\infty {e^{ - \alpha } \alpha ^n \,d\alpha } = \frac{{( - 1)^n n!}}<br />
{{(m + 1)^{n + 1} }}.

    --

    As for the first integral, set x=\frac1u.
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    \int_0^1 {x^m \ln ^n x\,dx} = \frac{{( - 1)^n }}<br />
{{(m + 1)^{n + 1} }}\int_0^\infty {e^{ - \alpha } \alpha ^n \,d\alpha } = \frac{{( - 1)^n n!}}<br />
{{(m + 1)^{n + 1} }}.
    I see you are begining to love the gamma function
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  7. #7
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    Aha, yes! It's really cool!
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  8. #8
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    Quote Originally Posted by Krizalid View Post
    Aha, yes! It's really cool!
    I have a love hate relationship with the Gamma function.
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  9. #9
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    Quote Originally Posted by Mess View Post
    Hi people!

    I can't solve these ones, they're tough!

    Solve \int {\frac{{\sqrt {\sqrt {x^4 + 1} - x^2 } }}<br />
{{x^4 + 1}}\,dx}

    Evaluate \int_0^1 {x\ln x\,dx}

    Thanks!
    And the answer (which may influence your decision to persist in working it out) is .............. :
    Attached Thumbnails Attached Thumbnails Integrals!-integral-2.gif  
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  10. #10
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    I wouldn't trust in the integrator, after substitution x=\frac1u leads another integral which can be solved easily.
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    I wouldn't trust in the integrator, after substitution x=\frac1u leads another integral which can be solved easily.
    The Integrator can be trusted, just not trusted to give the answer in its simplest form.

    -Dan
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  12. #12
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    Quote Originally Posted by topsquark View Post
    The Integrator can be trusted, just not trusted to give the answer in its simplest form.

    -Dan
    Mathematicians (the extremists) do not use calculator, computers, and machines.
    Because they shall say this is a direct violation against what math is about.
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