You would calculate the integral from 0 to 3 of sqrt((dx/dt)^2 + (dy/dt)^2)dt

dx/dt = [(5 + t)*1 - t*1]/(5 + t)^2 = [5 + t - t]/(5 + t)^2 = 5/(5 + t)^2

(dx/dt)^2 = 25/(5 + t)^4

dy/dt = 1/(5 + t)

(dy/dt)^2 = 1/(5 + t)^2

(dx/dt)^2 + (dy/dt)^2 = 25/(5 + t)^4 + 1/(5 + t)^2 = 25/(5 + t)^4 + (5 + t)^2/(5 + t)^4 = [25 + 25 + 10t + t^2]/(5 + t)^4 = [50 + 10t + t^2]/(5 + t)^4

sqrt((dx/dt)^2 + (dy/dt)^2) = sqrt([50 + 10t + t^2]/(5 + t)^4) = sqrt(t^2 + 10t + 50)/(t + 5)^2

Rewrite it as sqrt((t + 5)^2 +25)/(t + 5)^2.

We want to integrate this. Let u = t + 5. Then du = dt. The lower limit t = 0 becomes u = 5. The upper limit t = 3 becomes u = 8. The integral of that expression from 0 to 3 in terms of u is the integreal from 5 to 8 of

sqrt(u^2 + 25)/u^2.

Apply u-substitution again. Let u = 5tan(v). The expression in terms of v becomes

sqrt((5tan(v))^2 + 25)/(5tan(v))^2 = sqrt(25tan(v)^2 + 25)/(25tan(v)^2) = sqrt(25(tan(v)^2 + 1))/(25tan(v)^2) = 5sqrt(sec(v)^2)/25tan(v)^2 = sqrt(sec(v)^2)/5tan(v)^2 = |sec(v)|/5tan(v)^2.

Now when u = 5, 5tan(v) = 5 so tan(v) = 1. So v = pi/4 or 5pi/4. When u = 8, 5tan(v) = 8 so tan(v) = 8/5. So v = tan^-1(8/5) or v = tan^-1(8/5) + pi. I will go with v1 = pi/4 and v2 = tan^-1(8/5) so that |sec(v)| = sec(v).

Now du = 5sec^2(v)dv. The integral becomes the integral fro pi/4 to tan^-1(8/5) of

sec(v)*5sec^2(v)/5tan^2(v) dv = 5sec^3(v)/5tan^2(v) dv

Use the fact that sec(v) = 1/cos(v) and tan(v) = sin(v)/cos(v). The integral becomes

5(1/cos(v))^3 / [5sin^2(v)/cos^2(v)] dv = {1/cos^3(v)} / {sin^2(v)/cos^2(v)} dv = {cos^2(v)/cos^3(v)} / {sin^2(v)} = {1/cos(v)} / {sin^2(v)} dv = 1 / {cos(v) sin^2(v)} dv.

Integrating this last expression is tricky. You want to rewrite 1 / {cos(v) sin^2(v)} as {sin^2(v) + cos^2(v)} / {cos(v) sin^2(v)} which simplifies to sec(v) + cos(v)/sin^2(v). I'm assuming you know the trig identity sin^2(v) + cos^2(v) = 1.

Thus the integral from pi/4 to tan^-1(8/5) of 1 / {cos(v) sin^2(v)} dv is equal to the integral from pi/4 to tan^-1(8/5) of

sec(v) + cos(v)/sin^2(v) dv = ln|sec(v) + tan(v)| -1/sin(v) evaluated from pi/4 to tan^-1(8/5) =

ln|sec(tan^-1(8/5)) + tan(tan^-1(8/5))| - 1/sin(tan^-1(8/5)) - { ln|sec(pi/4) + tan(pi/4)| - 1/sin(pi/4) }.

This is very messy, but for my final answer I get

ln(sqrt(89)/5 + 8/5) - ln(sqrt(2) + 1) - sqrt(89)/8 + sqrt(2).