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    ÊÇÑíÎ ÇáÊÓÌíá
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    áÊäÔíØ ÇáÐåä 3


  2. Top | #2

    ÊÇÑíÎ ÇáÊÓÌíá
    Jun 2013
    ÑÞã ÇáÚÖæíÉ
    16609
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    ááÊÐßíÑ ÝÞØ¡ ãÚ ÇáÔßÑ

  3. Top | #3

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    ÑÏ: áÊäÔíØ ÇáÐåä 3

    $\LARGE {\color{Red} (1)\int (2x+3)lnx dx;;u=lnx\rightarrow du=\frac{1}{x}dx} $
    ${\color{Red} dv=2x+3\rightarrow v=x^{2}+3x}$
    ${\color{Red} I=(x^{2}+3x)lnx-\int (x^{2}+3x)\frac{1}{x}=(x^{2}+3x)lnx-\frac{x^{2}}{2}+3x+c}$

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     (07-09-2016)

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    ÊÇÑíÎ ÇáÊÓÌíá
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    ÑÏ: áÊäÔíØ ÇáÐåä 3

    $\LARGE {\color{DarkBlue} (2)\int (2x-1)e^{2x+3}dx;;u=2x-1\rightarrow du=2dx }$
    $ {\color{DarkBlue} dv=e^{2x+3}\rightarrow v=1/2e^{2x+3}} $
    ${\color{DarkBlue} I=1/2(2x-1)e^{2x+3}-\int e^{2x+3}=(1/2)(2x-1)e^{2x+3}-(1/2)e^{2x+3}+c}$

  6. The Following User Says Thank You to Ìáíá ÇáÒíÇÏí For This Useful Post:

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