$\text{ If }\int \frac{{\tan }^{-1}x}{x^4}dx=A\frac{{\tan }^{-1}x}{x^3}+B\log \left|\frac{x^2+1}{x^2}\right|+\frac{c}{x^2}+k\text{ then }A-B+C=$

$\int \frac{{\tan }^{-1}x}{x^4}dx;u={\tan }^{-1}xv=\frac{1}{x^4}$
$= \left({\tan }^{-1}\mathrm{x}\right)\left(\frac{{\mathrm{x}}^{-3}}{-3}\right)-\int \frac{1}{(1+{\mathrm{x}}^2)}\left(\frac{{\mathrm{x}}^{-3}}{-3}\right)$
$$\begin{gathered} = \frac{-{\tan }^{-1}\mathrm{x}}{3{\mathrm{x}}^3}+\frac{1}{3}\int \frac{1}{{\mathrm{x}}^3(1+{\mathrm{x}}^2)}\mathrm{dx} \\ = \frac{-{\tan }^{-1}\mathrm{x}}{3{\mathrm{x}}^3}+\frac{1}{3}\int [\frac{-1}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^3}+\frac{\mathrm{x}}{1+{\mathrm{x}}^2}]\mathrm{dx} \\ = \frac{-{\tan }^{-1}\mathrm{x}}{3{\mathrm{x}}^3}+\frac{1}{3}[-\mathrm{logx}-\frac{1}{2{\mathrm{x}}^2}+\frac{1}{2}\log (1+{\mathrm{x}}^2\left)\right]+\mathrm{c} \\ = \frac{-{\tan }^{-1}\mathrm{x}}{3{\mathrm{x}}^3}+\frac{1}{3}[\frac{-1}{2{\mathrm{x}}^2}+\frac{1}{2}\log (\frac{1+{\mathrm{x}}^2}{{\mathrm{x}}^2}\left)\right]+\mathrm{c}, \\ \Rightarrow \mathrm{A} = -\frac{1}{3}, \mathrm{B} = \frac{1}{6}, \mathrm{C} = \frac{-1}{6} \\ \Rightarrow \mathrm{A}-\mathrm{B}+\mathrm{C} = \frac{-2}{3} \end{gathered}$$