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

You just need to expand the square:
$$
\int_{4}^{8}{\sqrt{1+\left(\frac{x^2-4}{4x}\right)^2}dx}=\int_{4}^{8}{\sqrt{1+\frac{x^4-8x^2+16}{16x^2}}dx}=\int_{4}^{8}{\sqrt{\frac{x^4+8x^2+16}{16x^2}}dx}=\int_{4}^{8}{\sqrt{\left(\frac{x^2+4}{4x}\right)^2}dx}=\int_{4}^{8}{\frac{x^2+4}{4x}dx}=\int_{4}^{8}{\frac{x}{4}+\frac{1}{x}dx}=\frac{8^2}{8}-\frac{4^2}{8}+\ln(8)-\ln(4)=6+\ln(2)
$$
