Displacement via Net Change Theorem

Let $u = t+1$, so $du = dt$. When $t=0$, $u=1$, and when $t=10$, $u=11$. The integral becomes $\int_1^{11} \frac{1}{u}\,du.$

This evaluates to $\ln|u|\Big|_1^{11} = \ln(11) - \ln(1) = \ln(11)$. Since velocity is positive throughout $[0,10]$, this value also represents the total distance traveled, and the displacement is $\ln(11)$ meters.

They are equal. Since v(t) = 1/(t+1) is positive for all t in [0,10], the object never reverses direction, so the total distance traveled matches the net displacement exactly, both equal to ln(11) meters.