For the parametric equation
x - 1/t^2,quad y = 1/t
we have
dx = -2/t^3 dt, quad dy= -1/t^2 dt
Thus, the infinitesimal arc length is
ds = sqrt{dx^2+dy^2}=sqrt{4/t^6 +1/t^4}dt = sqrt{4+t^2} dt/t^3
Thus, the required arc length is
int_1^2 sqrt{4+t^2} dt/t^3
To evaluate this, lets try the substitution 4+t^2 = u^2. Then 2tdt = 2 u du. We also have u = sqrt{5} and u=sqrt8 at t=1 and t=2, respectively.
Thus the arc length is
int_{sqrt5}^{sqrt{8}} u (udu)/(u^2-4)^2
We carry out the partial fractions expansion
u^2/(u^2-4)^2 = A/(u-2)+B/(u+2)+C/(u-2)^2+D/(u+2)^2
This leads to
u^2 = A(u-2)(u+2)^2+B(u-2)^2(u+2)+C(u+2)^2+D(u-2)^2
Substituting u=2 and u=-2, respectively, gives 2^2 = 16C and 2^2=16D, leading to
C = D = 1/4
Substituting this back leads to
u^2 = A(u-2)(u+2)^2+B(u-2)^2(u+2)+1/4((u+2)^2+(u-2)^2)
qquad = A(u-2)(u+2)^2+B(u-2)^2(u+2)+1/2(u^2+4)
and so
1/2(u^2-4) = A(u-2)(u+2)^2+B(u-2)^2(u+2) implies
1/2 = A(u+2)+B(u-2)
Substituting u=pm 2 in the above leads to
A = 1/8, quad B = -1/8
So,
u^2/(u^2-4)^2 = 1/8 1/(u-2)-1/8 1/(u+2)+1/4 1/(u-2)^2+1/4 1/(u+2)^2
and the arc length is
int_{sqrt5}^{sqrt{8}} (1/8 1/(u-2)-1/8 1/(u+2)-1/4 1/(u-2)^2-1/4 1/(u+2)^2) du
qquad = (1/8 ln|(u-2)/(u+2)|-1/4 1/(u-2) -1/4 1/(u+2) ) _{sqrt5}^{sqrt{8}}
qquad = (1/8 ln|(u-2)/(u+2)|-1/2 u/(u^2-4)) _{sqrt5}^{sqrt{8}} ~~ 0.905
