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Adrianx
user (0)
Pe: 2008-12-07T11:16:57+02:00 In: In:

Integrala

Prin ce metoda se rezolva integrala de mai jos? Imi puteti da niste idei cum as putea sa o rezolv? Multumesc anticipat.

    \[ \int {\frac{{dx}}{{1 + \sqrt {x - 1} }}} \]

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4 raspunsuri

  1. ali
    maestru (V)

    Indicatie:inmulteste cu conjugatul.

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  2. Adrianx
    user (0)

        \[ \int {\frac{{dx}}{{1 + \sqrt {x - 1} }} = \int {\frac{{(1 - \sqrt {x - 1} )}}{{(1 + \sqrt {x - 1} )(1 - \sqrt {x - 1} )}}dx} } = \int {\frac{{1 - \sqrt {x - 1} }}{{1 - x + 1}}} dx = \[ \int {\frac{{1 - \sqrt {x - 1} }}{{2 - x}}dx = \int {\frac{1}{{2 - x}}dx - \int {\frac{{\sqrt {x - 1} }}{{2 - x}}dx} } } \] \]

    de aici m-am blocat.

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  4. gigi.becali
    user (0)

    notezi \sqrt{x-1}=t \Rightarrow x=t^2 + 1 \Rightarrow dx = 2tdt \Rightarrow I=2\int{\frac{t}{1+t}dt} care e f simplu de calculat

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  5. ali
    maestru (V)

    Adrianx wrote: de aici m-am blocat.


    Banuiesc ca la a 2 integrala te-ai blocat deoarece prima este destul de simpla.Continuare:

    *** QuickLaTeX cannot compile formula:

	\[
	\begin{array}{l}
	 \sqrt {x - 1}  = t \Rightarrow x = t^2  + 1 \Rightarrow dx = 2tdt \\
	 revenim{\rm  la integrala]
	

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\begin{array} on input line 10 ended by \end{document}.
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Missing } inserted.
leading text: \end{document}
Emergency stop.
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