気付かないと解けないかも

\begin{align*} \int_{0}^{\infty}\frac{1}{\left(1+x\right)\left(a^{2}+\log^{2}x\right)}dx & =\int_{\infty}^{0}\frac{1}{\left(1+\frac{1}{y}\right)\left(a^{2}+\log^{2}x\right)}\left(-\frac{1}{y^{2}}\right)dy\cmt{y=\frac{1}{x}}\\ & =\int_{0}^{\infty}\frac{\frac{1}{y}}{\left(y+1\right)\left(a^{2}+\log^{2}x\right)}dy\\ & =\int_{0}^{\infty}\frac{1+\frac{1}{y}-1}{\left(y+1\right)\left(a^{2}+\log^{2}x\right)}dy\\ & =\int_{0}^{\infty}\frac{1+\frac{1}{y}}{\left(y+1\right)\left(a^{2}+\log^{2}x\right)}dy-\int_{0}^{\infty}\frac{1}{\left(y+1\right)\left(a^{2}+\log^{2}x\right)}dy\\ & =\frac{1}{2}\int_{0}^{\infty}\frac{1+\frac{1}{y}}{\left(y+1\right)\left(a^{2}+\log^{2}x\right)}dy\\ & =\frac{1}{2}\int_{0}^{\infty}\frac{y+1}{y\left(y+1\right)\left(a^{2}+\log^{2}x\right)}dy\\ & =\frac{1}{2}\int_{0}^{\infty}\frac{1}{y\left(a^{2}+\log^{2}x\right)}dy\\ & =\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{a^{2}+\log^{2}y}d\log y\\ & =\frac{1}{2}\left[\frac{1}{a}\tan^{\bullet}\left(\frac{\log y}{a}\right)\right]_{\log y\rightarrow-\infty}^{\log y\rightarrow\infty}\\ & =\frac{1}{2a}\left(\frac{\pi}{2}\sgn\left(a\right)-\left(-\frac{\pi}{2}\sgn\left(a\right)\right)\right)\\ & =\frac{\pi}{2a}\sgn\left(a\right)\\ & =\frac{\pi}{2\left|a\right|} \end{align*}