E-Mag Biot-Savart law (line current - Magnetic field)
For each infinitesimal segment of the wire, \$\text{d}L\$, this expression holds: $$\vec{\text{d}H} =-\frac{1}{4 \pi}\frac{I\:x\:\text{d}y}{\left(x^2+y^2\right)^\frac32}\left&lt;0,0,1\right&gt;$$

The above follows from applying a cross-product to the distance \$r\$ for any given infinitesimal segment.

\$x_0=2\$ is the closest approach. \$y_0=3\$.

$$\begin{align*}\frac{I\: x_0}{4\pi}\int_{y=0}^{y=\infty}\frac{\text{d}y}{\left[x_0^2+\left(y-y_0\right)^2\right]^\frac32}&amp;=\frac{I\:x_0}{4\pi}\int_{y=-3}^{y=\infty}\frac{\text{d}y}{\left(x_0^2+y^2\right)^\frac32}\\\\&amp;=\frac{I\: x_0}{4\pi}\left[\frac{y}{x_0^2\sqrt{x_0^2+y^2}}\right]_{y=-3}^{y=\infty}\\\\&amp;=\frac{I}{4\pi}\left[\frac{1}{x_0}-\frac{-3}{x_0\sqrt{x_0^2+\left(-3\right)^2}}\right]\\\\&amp;=\frac{I}{4\pi}\left[\frac{1}{2}+\frac{3}{2\sqrt{13}}\right]\\\\&amp;=\frac{1}{4\pi}\left[1+\frac{3}{\sqrt{13}}\right]\end{align*}$$

Note the minus signs cancel out in this definite integral.

Another way of saying this is that the short, additional length of wire adds to the net magnetic strength. It doesn't subtract from it.