Calculate the length of the curve: $y = \frac{1}{x}$ between points $(1,1)$ and $(2, \frac{1}{2})$.
What I tried: $$ \int_a^b\sqrt{(x')^2+(y')^2} dt$$ $$r(t) =(t,1/t) $$ $$\int_1^2\sqrt{(1)^2+\left(\frac{1}{t^2}\right)^2}\,dt$$ $$\int_1^2\sqrt{1+\frac{1}{t^{4}}}\,dt$$
However, if my procedure to here is correct (I am not sure), then I wanted to solve this integral and that would give me my solution.
$$\int_1^2 \frac{1}{t^2} \sqrt{t^4+1}\,dt$$ However, I do not know what substitution to make in this integral for this to work.
Your derivation of the integral is perfectly fine, but to actually evaluate it requires elliptic integrals. We can rewrite the integral as $$\int_1^2\frac{t^2+1/t^2}{\sqrt{t^4+1}}\,dt$$ Then using the formulas on page 7 of my Elliptic Integrals and Functions monograph transforms this into $$\int_{F(\pi/2,1/2)}^{F(2\tan^{-1}2,1/2)}\left(\frac2{\operatorname{sn}^2(u,1/2)}-1\right)\,du$$ where Mathematica convention is used. Page 8 of the monograph (the $B_k$ part) reduces this to $f(2)-f(1)$ where $$f(p)=F(2\tan^{-1}p,1/2)-2E(2\tan^{-1}p,1/2)+\frac{(p^2-1)\sqrt{p^4+1}}{p(p^2+1)}$$ The addition formulae on page 4 reduce this even further, to $$F\left(\sin^{-1}\frac3{\sqrt{17}},\frac12\right)-2E\left(\sin^{-1}\frac3{\sqrt{17}},\frac12\right)+\frac{15}{2\sqrt{17}}$$