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integration FP3

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Hi everyone

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


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Reply 1
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?
Reply 2
Original post by M14B
Complete the square. Does this help?


I have tried, it doesn't seem helpful though.


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Reply 3
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

Unparseable latex formula:

\displaystyle[br]\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+18x + 13 = 8x + 12 + 1
(edited 8 years ago)
Reply 4
Original post by Zacken
Split it into

Unparseable latex formula:

\displaystyle[br]\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+18x + 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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Reply 5
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. dxx2±a2\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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