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gabriell
Pe: 2013-03-14T17:28:46+02:00 In: In:

Integrale

Am dat peste cateva integrale care nu reusesc sa le rezolv,am incercat cu mai multe substitutii insa nu s-au simplificat integralele si ajungeam la structuri asemanatoare cu prima integrala,probabil se rezolva cu ceva substitutii de care eu inca nu stiu 🙂

    $$\int\limits_0^1 {\frac{{\sqrt x }}{{(x + 3)\sqrt {x + 3} }}} dx$$

    $$\int\limits_0^3 {\frac{{\sqrt x }}{{\sqrt x + \sqrt {3 - x} }}} dx$$

    $$\int\limits_0^{(\sqrt 3 + \sqrt 5 )/\sqrt 2 } {\frac{{{x^2} + 1}}{{{x^4} + 1}}} dx$$

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3 raspunsuri

  1. DD
    profesor

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  2. gabriell

    la ex 3 am gasit A=C=0 si B=D=1/2,pe urma

        $${x^2} \pm \sqrt 2 x + 1;\Delta \langle 0$$

        $$\frac{1}{2}\int {\frac{{dx}}{{{{\left( {x + \frac{{\sqrt 2 }}{2}} \right)}^2} + {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}} + \frac{1}{2}\int {\frac{{dx}}{{{{\left( {x - \frac{{\sqrt 2 }}{2}} \right)}^2} + {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}} $$

        $$\frac{{\sqrt 2 }}{2}\left[ {\left( {arctg\sqrt 2 \left( {x + \frac{{\sqrt 2 }}{2}} \right)} \right) + arctg\sqrt 2 \left( {x + \frac{{\sqrt 2 }}{2}} \right)} \right]_0^{\frac{{\sqrt 3 + \sqrt 5 }}{{\sqrt 2 }}}$$

        $$\frac{{\sqrt 2 }}{2}\left[ {arctg(\sqrt 3 + \sqrt 5 - 1) + \frac{\pi }{4} + arctg(\sqrt 3 + \sqrt 5 + 1) - \frac{\pi }{4}} \right]$$

    aici nu mai stiu ce pot sa fac cu arctg

    la ex 2 am obtinut A=C=-1/4 B=1/4 D=E=1/2

        $$ - \frac{1}{4}\int {\frac{{dt}}{{1 + t}}} + \frac{1}{4}\int {\frac{{t - 1}}{{{t^2} + 1}}} dt + \frac{1}{2}\int {\frac{{t + 1}}{{{{\left( {{t^2} + 1} \right)}^2}}}} dt$$

    dupa calcule am obtinut expresia

        $$ - \frac{1}{4}\ln \left| {1 + t} \right| + \frac{1}{4}\left( {\frac{1}{2}\ln \left| {1 + {t^2}} \right| - arctgt} \right) + \frac{1}{2}\left[ { - \frac{1}{2}.\frac{1}{{{t^2} + 1}} + arctgt + \frac{1}{2}\left( {\frac{t}{{{t^2} + 1}} - arctgt} \right)} \right]$$

    si intr-un final am ajuns aici

        $$I = - 6\left( { - \frac{1}{4}\ln \left| {1 + t} \right| + \frac{1}{8}\ln \left| {1 + {t^2}} \right| - \frac{1}{{4({t^2} + 1)}} + \frac{t}{{4({t^2} + 1)}}\left| {_{ - \infty }^0} \right.} \right)$$

  4. DD
    profesor

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