= Solution
Parametrize the segment by $\mathbf r(t)=(1-t,t,0)$, $0\leq t\leq1$, so $\mathbf r'=(-1,1,0)$. For the first field, $\mathbf F_1=(-t,1-t,-1+t)$ and its <dot product> with the tangent is one. Therefore
$$
\boxed{\int_{\rm segment}\mathbf F_1\cdot d\mathbf r=1}.
$$
For the second field, use the <fundamental theorem for line integrals>. The endpoints both have $r=1$, so their potential values agree and
$$
\boxed{\int_{\rm segment}\mathbf F_2\cdot d\mathbf r=0}.
$$
Directly, its parameter integrand is $(2t-1)/[1+(1-t)^2+t^2]^{3/2}$, which changes sign under $t\mapsto1-t$ and integrates to zero.
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