Originally Posted by

**funnytim** Hey everyone,

I'm needing to take the integral of the following. However I don't get the correct answer (when I put in the bounds), so I'm wondering if I integrated it wrong somewhere?

The equation is

$\displaystyle \int_2^3 \sqrt{1 + (x^2/8 -2/x^2)^2}\ dx$

So I started off by multiplying out the square,

$\displaystyle \int_2^3 \left(\sqrt{1 + x^4/64 - 2x^2/8x^2 - 2 x^2/8x^2 + 4/x^4}\right)\ dx$

=

$\displaystyle \int_2^3 \left(\sqrt{1 + x^4/64 - 1/2 + 4/x^4}\right)\ dx$

Square rooted everything:

$\displaystyle \int_2^3 \left({1 + x^3/8 - 1/\sqrt{2} + 2x^{-3}}\right)\ dx$ ** this is incorrect**

Integrated it: $\displaystyle x +x^4/32 - x/ \sqrt{2} - x^{-2}$

Did I do something wrong up there?

Thanks in advance!