integral (square root)

• Sep 10th 2010, 07:23 PM
mybrohshi5
integral (square root)
I am having trouble solving this integral
$\displaystyle \int ^2_1 \sqrt(4x^2+4+\frac{1}{x^2}) dx$

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 :)
• Sep 10th 2010, 07:41 PM
TheEmptySet
Quote:

Originally Posted by mybrohshi5
I am having trouble solving this integral
$\displaystyle \int ^2_1 \sqrt(4x^2+4+\frac{1}{x^2}) dx$

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 :)

$\displaystyle 4x^2+4+\frac{1}{x^2}=\left(2x+\frac{1}{x} \right)\left(2x+\frac{1}{x} \right)$
• Sep 10th 2010, 07:43 PM
Soroban
Hello, mybrohshi5!

Quote:

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

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

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

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

Got it?

• Sep 10th 2010, 07:49 PM
mybrohshi5
Oh boy i feel stupid that i missed that haha.

Thanks for both of the fast replies :D