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Math Help - Integration

  1. #1
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    Integration

    ∫x^2/√(x^2 + 1) dx

    How can I do this?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by geton View Post
    ∫x^2/√(x^2 + 1) dx

    How can I do this?
    Integration by parts........
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  3. #3
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    Quote Originally Posted by geton View Post
    ∫x^2/√(x^2 + 1) dx

    How can I do this?
    <br />
\int {\frac{{x^2 }}<br />
{{x^2  + 1}}\;dx = } \int {\frac{{x^2  + 1 - 1}}<br />
{{x^2  + 1}}\;dx}  = \int {1 - \frac{1}<br />
{{x^2  + 1}}\;dx}  \ldots


    edit: oops...missed your \sqrt{}...you need integration by parts as "mathstud" showed.
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  4. #4
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    Yes I got it. Thanks.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xifentoozlerix View Post
    <br />
\int {\frac{{x^2 }}<br />
{{x^2  + 1}}\;dx = } \int {\frac{{x^2  + 1 - 1}}<br />
{{x^2  + 1}}\;dx}  = \int {1 - \frac{1}<br />
{{x^2  + 1}}\;dx}  \ldots


    edit: oops...missed your \sqrt{}...you need integration by parts as "mathstud" showed.
    did you notice there is a square root in the denominator?

    ok, i just saw your edit
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    alternatively, we could make a substitution, u^2 = x^2 + 1

    or do a trig substitution, x = \tan \theta
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    alternatively, we could make a substitution, u^2 = x^2 + 1
    Trig. sub. will do it, but I'm afraid that your first sub. won't work.

    Or make \frac{{x^2 }}<br />
{{\sqrt {x^2  + 1} }} = \sqrt {x^2  + 1}  - \frac{1}<br />
{{\sqrt {x^2  + 1} }}.

    You can tackle the first one via trig. - hyperbolic substitution or integration by parts, the another one is a classic one.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Trig. sub. will do it, but I'm afraid that your first sub. won't work.

    Or make \frac{{x^2 }}<br />
{{\sqrt {x^2  + 1} }} = \sqrt {x^2  + 1}  - \frac{1}<br />
{{\sqrt {x^2  + 1} }}.

    You can tackle the first one via trig. - hyperbolic substitution or integration by parts, the another one is a classic one.
    hehe, i figure there might have been a problem with the first substitution, but i was too lazy to check. that's why i also put the trig sub there (i hoped to avoid trig sub altogether--you're rubbing off on me )
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