= Solution
Use the branch from part (b) and close the real interval $[-R,R]$ with the upper semicircle, oriented counterclockwise. For $\operatorname{Im}z\geq0$, the number $z+i\alpha$ has positive imaginary part, so the numerator is a <holomorphic logarithm> throughout a neighbourhood of the contour and its interior. The branch point $-i\alpha$ lies strictly below the contour. The only enclosed <pole> of
$$
H(z)=\frac{\operatorname{Log}(z+i\alpha)}{z^2+\beta^2}
$$
is the <simple pole> at $i\beta$, with <residue>
$$
\operatorname{Res}_{z=i\beta}H=\frac{\operatorname{Log}(i(\alpha+\beta))}{2i\beta}
=\frac{\log(\alpha+\beta)+i\pi/2}{2i\beta}.
$$
To justify discarding the arc, take $R>\max\{\beta,\alpha+1\}$. On it, $|z+i\alpha|$ lies between $R-\alpha$ and $R+\alpha$, and the argument has bounded absolute value. Hence
$$
\left|\int_{\text{arc}}H(z)\,dz\right|
\leq\frac{\pi R\{\log(R+\alpha)+3\pi/2\}}{R^2-\beta^2}\longrightarrow0.
$$
The real-axis integral is absolutely convergent, since its integrand is $O(\log|x|/x^2)$ at infinity and has no real singularities. The <residue theorem> therefore gives
$$
\boxed{\int_{-\infty}^{\infty}\frac{\operatorname{Log}(x+i\alpha)}{x^2+\beta^2}\,dx
=\frac\pi\beta\log(\alpha+\beta)+\frac{i\pi^2}{2\beta}.}
$$
Finally, $\operatorname{Re}\operatorname{Log}(x+i\alpha)=\tfrac12\log(x^2+\alpha^2)$. Taking twice the real part yields
$$
\boxed{\int_{-\infty}^{\infty}\frac{\log(x^2+\alpha^2)}{x^2+\beta^2}\,dx
=\frac{2\pi}{\beta}\log(\alpha+\beta).}
$$
The branch choice affects the first integral's imaginary part; the second integral is real and independent of that choice.
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