bend measurement and calculating $\int_4^8 \sqrt{1+{\left(\frac{{x^2-4}}{4x}\right)^2}} $ How can i get the measure of this bend : $y=\left(\frac{x^2}{8}\right)-\ln(x)$ between $4\le x \le 8$. i solved that a bit according to the formula $\int_a^b \sqrt{1+{{f'}^2}} $:$$\int_4^8 \sqrt{1+{\left(\frac{x^2-4}{4x}\right)^2}} $$ $$= \cdots$$I don't know how I calculate this integration.the answer is: 6+ln2

HINT:
$$\sqrt{1+\left(\frac{x^2-4}{4x}\right)^2}=\frac{\sqrt{16x^2+(x^4-8x^2+16)}}{4x}=\frac{x^2+4}{4x}$$
