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Math Help - Integrals = no mercy

  1. #1
    Member iPod's Avatar
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    Integrals = no mercy

    5) (i) Use the substitution u^2=2x+1 to show that, x>4

    \int\frac{3}{(x-4)\sqrt{2x+1}}dx=\ln(\frac{\sqrt{2x+1}-3}{\sqrt{2x+1}+3})+K

    where K is a constant.

    (ii) Show that (from ln 8 to ln 3) \int\frac{2}{e^x\sqrt{e^x+1}}=\frac{7}{12}+\ln\fra  c{2}{3}

    :'( hints please
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  2. #2
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    Quote Originally Posted by iPod View Post
    (ii) Show that (from ln 8 to ln 3) \int\frac{2}{e^x\sqrt{e^x+1}}=\frac{7}{12}+\ln\fra  c{2}{3}

    :'( hints please
    For

     \int_{ln(3)}^{ln(8)}\frac{2}{e^x\sqrt{e^x+1}}dx

    consider u = e^x+1 \Rightarrow \frac{du}{dx}= e^x
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  3. #3
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    Quote Originally Posted by iPod View Post

    5) (i) Use the substitution u^2=2x+1 to show that, x>4

    \int\frac{3}{(x-4)\sqrt{2x+1}}dx=\ln(\frac{\sqrt{2x+1}-3}{\sqrt{2x+1}+3})+K
    Hint is given, I think you can show what you've done.
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  4. #4
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    Quote Originally Posted by iPod View Post
    5) (i) Use the substitution u^2=2x+1 to show that, x>4

    \int\frac{3}{(x-4)\sqrt{2x+1}}dx=\ln(\frac{\sqrt{2x+1}-3}{\sqrt{2x+1}+3})+K

    where K is a constant.

    (ii) Show that (from ln 8 to ln 3) \int\frac{2}{e^x\sqrt{e^x+1}}=\frac{7}{12}+\ln\fra  c{2}{3}

    :'( hints please
    For the first, you have to let u = \sqrt{2x + 1}.

    This means that \frac{du}{dx} = \frac{1}{\sqrt{2x + 1}} = \frac{1}{u}.


    With some algebraic manipulation, you can see that

    u^2 = 2x + 1

    x = \frac{1}{2}u^2 - \frac{1}{2}


    Therefore x - 4 = \frac{1}{2}u^2 - \frac{9}{2}.


    So \frac{3}{(x - 4)\sqrt{2x + 1}} = \frac{3}{x - 4}\,\frac{1}{\sqrt{2x + 1}}

     = \frac{3}{\frac{1}{2}u^2 - \frac{9}{2}} \, \frac{du}{dx}

     = \frac{6}{u^2 - 9}\,\frac{du}{dx}

     = \frac{6}{(u - 3)(u + 3)}\,\frac{du}{dx}.


    Can you go from here? You need to use partial fractions.
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