If the loop of wire is square and the second wire with a current passes directly through the center of the wire, then everything is symmetric. There is some $\vec{B}$ which you could find - it has magnitude $\frac{\mu_0 I}{2\pi r}$ in the right-handed direction.
Lets imagine you were being silly and ignored the symmetry of the problem. Instead you calculated every little detail of the problem, laboriously doing the integrals.
So, imagine point "A" is at $(-a,a)$ and point "B" is at $(a,a)$, "a" being the size of the square. The wire is at $(0,0)$ going into the plane. Then $$\vec{B}=\frac{\mu_0 I}{2\pi r}(\sin(\theta),-\cos{\theta})=\frac{\mu_0 I}{2\pi \sqrt{x^2+y^2}}(\frac{y}{\sqrt{x^2+y^2}},-\frac{x}{\sqrt{x^2+y^2}})$$
(The direction $(\sin(\theta),-\cos{\theta})$ is perpendicular to the radial vector $(\cos(\theta),\sin(\theta))$ in the clockwise sense. And I write out $\cos(\theta)=\frac{x}{r}$, $\sin(\theta)=\frac{y}{r}$, and $r=\sqrt{x^2+y^2}$)
We also have, along the path I choose, $d\vec{l}=(dx,0)$. Substitute $y=a$. Then...
$$\int_A^B \vec{B}\cdot d\vec{l}=\int_{-a}^a\frac{\mu_0 I}{2\pi \sqrt{x^2+a^2}}\frac{a dx}{\sqrt{x^2+a^2}}$$
If I plug this into Mathematica, I find:
$$\int_{-a}^a\frac{a dx}{2\pi (x^2+a^2)}=\frac{1}{4}$$
but this should have been obvious from the start. If you were to write out all of the other integrals, $\int_B^C \vec{B}\cdot d\vec{l}$, $\int_C^D \vec{B}\cdot d\vec{l}$, $\int_D^A \vec{B}\cdot d\vec{l}$, you'd find that you're evaluating exactly the same quantity four times! So if we know the sum of the four quantities is $\mu_0 I$, and we know each quantity equals every other quantity, each integral was bound to be $\frac{1}{4}\mu_0 I$ in the first place. The integration was wasted effort!