Definite integral $\int_1^2 \sqrt{1+\left(-x^{-2}+x^2/4\right)^2}\,dx$ I'm having trouble solving this integral. It relates to an arc length question. I tried Wolfram|Alpha, but when it solves it doesn't give me the option to view step-by-step. 
Integral:
$$
\int_{1}^{2} \sqrt{1+\left(-x^{-2}+\frac{x^2}{4}\right)^2}\,dx
$$
Original question: 
$$
y=\frac{1}{x} + \frac{x^3}{12}
$$
Find the arc length of the function on [1,2]. 
I used the equation:
$$
L = \int_a^b \sqrt{1+(f'(x))^2}
 dx$$

If you expand the square, you can incorporate the $+1$ term into the expansion and then neatly factor everything into a quantity that has a simple square root. $$1+\left(-x^{-2}+\frac{x^2}{4}\right)^2 = 1+\left(x^{-4}-\frac{1}{2}+\frac{x^4}{16}\right) \\ = x^{-4}+\frac{1}{2}+\frac{x^4}{16}  \\ = \left(x^{-2}+\frac{x^2}{4}\right)^2$$ and hence $$
\int_{1}^{2} \sqrt{1+\left(-x^{-2}+\frac{x^2}{4}\right)^2}\,dx
 = 
\int_{1}^{2} \sqrt{\left(x^{-2}+\frac{x^2}{4}\right)^2}\,dx
$$
Can you take it from here?
