integral problem (1 Viewer)

Tye F · Jan 19, 2010

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referrelative="t" o:spt="75" coordsize="21600,21600">how does the integral of 2(2x+2)/(8-2x-x^2)^0.5</v:shapetype>
<v:shapetype stroked="f" filled="f" path="m@4@5l@4@11@9@11@9@5xe" o

referrelative="t" o:spt="75" coordsize="21600,21600">=-4(8-2x-x^2)^0.5</v:shapetype>

untouchablecuz · Jan 19, 2010

Tye F said:
<v:shapetype id="_x0000_t75" stroked="f" filled="f" path="m@4@5l@4@11@9@11@9@5xe" o="">how does the integral of 2(2x+2)/(8-2x-x^2)^0.5</v:shapetype><v:shapetype stroked="f" filled="f" path="m@4@5l@4@11@9@11@9@5xe" o="">=-4(8-2x-x^2)^0.5</v:shapetype>

$\\2\int \frac{2x+2}{\sqrt{8-2x-x^2}}\,dx\\=-2\int \frac{-2-2x}{\sqrt{8-2x-x^2}}\,dx\\=-2\int \frac{f'(x)}{\sqrt{f(x)}}\,dx, \text{ where }f(x)=8-2x-x^2\\=-2\int \frac{1}{\sqrt{f(x)}}\,d\left(f(x)\right)\\=-4\sqrt{f(x)}+C\\=-4\sqrt{8-2x-x^2}+C$

integral problem (1 Viewer)

Tye F

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untouchablecuz

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