The general solution of the differential equation \(\frac{1}{x} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\tan ^{-1} x\) is

\(\begin{aligned}
& \frac{1}{x} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\tan ^{-1} x \\
& \mathrm{~d} y=x \cdot \tan ^{-1} x \mathrm{~d} x
\end{aligned}\)
Integrating on both sides we get,
\(\begin{aligned}
& y=\int x \cdot \tan ^{-1} x \mathrm{~d} x \\
& y=\tan ^{-1} x \int x \mathrm{~d} x-\int\left(\frac{\mathrm{d}}{\mathrm{~d} x} \tan ^{-1} x \cdot \int x \mathrm{~d} x\right) \mathrm{d} x \\
& y=\tan ^{-1} x \cdot \frac{x^2}{2}-\int\left(\frac{1}{1+x^2} \times \frac{x^2}{2}\right) \mathrm{d} x \\
& y=\tan ^{-1} x \cdot \frac{x^2}{2}-\frac{1}{2} \int\left(\frac{x^2}{1+x^2}\right) \mathrm{d} x \\
& =\frac{x^2 \cdot \tan ^{-1} x}{2}-\frac{1}{2}\left(\int \frac{x^2+1}{x^2+1}-\int \frac{1}{1+x^2} \mathrm{~d} x\right) \\
& \Rightarrow y=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+\mathrm{c}
\end{aligned}\)