The particular solution of differential equation \(\left(1+y^2\right)(1+\log x) \mathrm{d} x+x \mathrm{~d} y=0\) at \(x=1, y=1\) is

\(\begin{aligned}
& \left(1+y^2\right)(1+\log x) \mathrm{d} x+x \mathrm{~d} y=0 \\
& \Rightarrow\left(1+y^2\right)(1+\log x) \mathrm{d} x=-x \mathrm{~d} y \\
& \Rightarrow\left(\frac{1+\log x}{x}\right) \mathrm{d} x=\left(\frac{-1}{1+y^2}\right) \mathrm{d} y
\end{aligned}\)
Integrating on both sides, we get
\(\Rightarrow \int \frac{(1+\log x)}{x} \mathrm{~d} x=-1 \int \frac{1}{1+y^2} \mathrm{~d} y\)
\(\Rightarrow \int \mathrm{tdt}=-\tan ^{-1} y+\mathrm{c} \quad \cdots\left[\begin{array}{l}\text {Let } 1+\log x=\mathrm{t} \\ \frac{1}{x} \mathrm{~d} x=\mathrm{dt}\end{array}\right]\)
\(\begin{aligned}
& \Rightarrow \frac{\mathrm{t}^2}{2}=-\tan ^{-1} y+\mathrm{c}...(i) \\
& \Rightarrow \frac{(1+\log x)^2}{2}=-\tan ^{-1} y+\mathrm{c}
\end{aligned}\)
\(\begin{aligned}
& \text { At } x=1, y=1 \\
& \Rightarrow \frac{(1+\log 1)^2}{2}=-\tan ^{-1}(1)+c \\
& \Rightarrow c=\frac{1}{2}+\frac{\pi}{4}
\end{aligned}\)
Substituting above value in (i), we get
\(\begin{aligned}
& \frac{(1+\log x)^2}{2}=-\tan ^{-1} y+\frac{1}{2}+\frac{\pi}{4} \\
& \frac{1}{2}+\log x+\frac{(\log x)^2}{2}=-\tan ^{-1} y+\frac{1}{2}+\frac{\pi}{4} \\
& \Rightarrow \log x+\frac{(\log x)^2}{2}+\tan ^{-1} y=\frac{\pi}{4}
\end{aligned}\)