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  1. Top | #1

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  3. Top | #3

    May 2014
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    $\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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    $\LARGE {\color{DarkBlue} (2)\int (2x-1)e^{2x+3}dx;;u=2x-1\rightarrow du=2dx }$
    $ {\color{DarkBlue} dv=e^{2x+3}\rightarrow v=1/2e^{2x+3}} $
    ${\color{DarkBlue} I=1/2(2x-1)e^{2x+3}-\int e^{2x+3}=(1/2)(2x-1)e^{2x+3}-(1/2)e^{2x+3}+c}$

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