integral (square root)

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September 10th 2010, 07:23 PM
mybrohshi5

integral (square root)

I am having trouble solving this integral
$\int ^2_1 \sqrt(4x^2+4+\frac{1}{x^2}) dx$

I think i can reduce the equation under the square root to an expression that is squared which would get rid of the square root (then it would be easy to evaluate from there), but i cannot seem to figure it out (by completing the square or any other way)

Thanks for any help :)
September 10th 2010, 07:41 PM
TheEmptySet

Quote:

Originally Posted by mybrohshi5

I am having trouble solving this integral
$\int ^2_1 \sqrt(4x^2+4+\frac{1}{x^2}) dx$

I think i can reduce the equation under the square root to an expression that is squared which would get rid of the square root (then it would be easy to evaluate from there), but i cannot seem to figure it out (by completing the square or any other way)

Thanks for any help :)

$4x^2+4+\frac{1}{x^2}=\left(2x+\frac{1}{x} \right)\left(2x+\frac{1}{x} \right)$
September 10th 2010, 07:43 PM
Soroban

Hello, mybrohshi5!

Quote:

$\displaystyle \int ^2_1 \sqrt{4x^2+4+\frac{1}{x^2}}\;dx$

Look at what's under the square root . . .

. . $4x^2 + 4 + \dfrac{1}{x^2} \;=\;\dfrac{4x^4 + 4x^2 + 1}{x^2} \;=\;\dfrac{(2x^2+1)^2}{x^2}$

Or you can factor it like this: . $4x^2 + 4 + \dfrac{1}{x^2} \;=\;\left(2x + \dfrac{1}{x}\right)^2$

Got it?
September 10th 2010, 07:49 PM
mybrohshi5

Oh boy i feel stupid that i missed that haha.

Thanks for both of the fast replies :D

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