Evaluating the arc length integral $\int\sqrt{1+\frac{x^4-8x^2+16}{16x^2}} dx$

Find length of the arc from $2$ to $8$ of $$y = \frac18(x^2-8 \ln x)$$

First I find the derivative, which is equal to $$\frac{x^2-4}{4x} .$$

Plug it into the arc length formula $$\int\sqrt{1+\left(\frac{dy}{dx}\right)^2} dx$$ and get

$$\int\sqrt{1+\frac{x^4-8x^2+16}{16x^2}} dx.$$

I am not sure how to proceed from here, as I cant figure out a way to put the sqrt argument into a form that I can find the square root of

• It seems you forgot to square the numerator of the derivative when substituting. Correcting this shows that the argument of $\sqrt{}$ is a perfect square. – Travis May 26 '16 at 8:01
• saw the mistake and edited it, I'm afraid I still don't see the solution – Joffrey Baratheon May 26 '16 at 8:04

$$1 + \frac{x^4 - 8 x^2 + 16}{16 x^2} = \frac{16 x^2}{16 x^2} + \frac{x^4 - 8 x^2 + 16}{16 x^2} =\frac{x^4 + 8 x^2 + 16}{16 x^2}$$ Can you write the rightmost expression as a square of another expression?