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#