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    Hi everyone

    Can anyone help me with Q11? I have no idea where to start. Thanks!


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    (Original post by 123321123)
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    Hi everyone

    Can anyone help me with Q11? I have no idea where to start. Thanks!


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    Complete the square. Does this help?
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    (Original post by M14B)
    Complete the square. Does this help?
    I have tried, it doesn't seem helpful though.


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    (Original post by 123321123)
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    Hi everyone

    Can anyone help me with Q11? I have no idea where to start. Thanks!


    Posted from TSR Mobile
    Split it into

    \displaystyle

\begin{equation*}\int_{\frac{-1}{4}}^{\frac{1}{6}} \frac{8x + 12}{\sqrt{4x^2 + 12x + 5}} \, \mathrm{d}x + \int_{\frac{-1}{4}}^{\frac{1}{6}}\frac{1 \, \mathrm{d}x}{\sqrt{4x^2 + 12x + 5}} = \int_{\frac{-1}{4}}^{\frac{1}{6}} \frac{(4x^2 + 12x+5)'}{\sqrt{4x^2 + 12x+5}} \, \mathrm{d}x + \int_{\frac{-1}{4}}^{\frac{1}{6}} \frac{1}{\sqrt{4x^2 + 12x + 5}}} \, \mathrm{d}x \end{equation*}

    Since 8x + 13 = 8x + 12 + 1
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    (Original post by Zacken)
    Split it into

    \displaystyle

\begin{equation*}\int_{\frac{-1}{4}}^{\frac{1}{6}} \frac{8x + 12}{\sqrt{4x^2 + 12x + 5}} \, \mathrm{d}x + \int_{\frac{-1}{4}}^{\frac{1}{6}}\frac{1 \, \mathrm{d}x}{\sqrt{4x^2 + 12x + 5}} = \int_{\frac{-1}{4}}^{\frac{1}{6}} \frac{(4x^2 + 12x+5)'}{\sqrt{4x^2 + 12x+5}} \, \mathrm{d}x + \int_{\frac{-1}{4}}^{\frac{1}{6}} \frac{1}{\sqrt{4x^2 + 12x + 5}}} \, \mathrm{d}x \end{equation*}

    Since 8x + 13 = 8x + 12 + 1
    Hi thanks for ur help.
    Could you explain further about how to integrate the second part ? there is a square root which confuses me.


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    (Original post by 123321123)
    Hi thanks for ur help.
    Could you explain further about how to integrate the second part ? there is a square root which confuses me.


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    Look in your formula booklet. \int \frac{\mathrm{d}x}{\sqrt{x^2 \pm a^2}} is a standard integral, so just complete the square as usual.
 
 
 
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