Answer
$$\int\frac{\tan^{-1}x}{x^2}dx=\frac{1}{2}\ln\frac{x^2}{x^2+1}-\frac{\tan^{-1}x}{x}+C$$
Work Step by Step
$$A=\int\frac{\tan^{-1}x}{x^2}dx=\int x^{-2}\tan^{-1}xdx$$
Use Formula 108, which states that
$$\int x^n\tan^{-1}axdx=\frac{x^{n+1}}{n+1}\tan^{-1}ax-\frac{a}{n+1}\int\frac{x^{n+1}dx}{1+a^2x^2}$$
with $n=-2$ and $a=1$:
$$A=\frac{x^{-1}}{-1}\tan^{-1}x-\frac{1}{-1}\int\frac{x^{-1}dx}{1+1^2x^2}$$ $$A=-\frac{\tan^{-1}x}{x}+\int\frac{dx}{x(1+x^2)}$$ $$A=-\frac{\tan^{-1}x}{x}+I$$
*Consider I:
$$I=\int\frac{dx}{x(1+x^2)}$$
Multiply both numerator and denominator by $x$: $$I=\int\frac{xdx}{x^2(1+x^2)}$$ $$I=\frac{1}{2}\int\frac{d(x^2)}{x^2(1+x^2)}$$
Set $u=x^2$
$$I=\frac{1}{2}\int\frac{du}{u(u+1)}=\frac{1}{2}\int\frac{u+1-u}{u(u+1)}du$$ $$I=\frac{1}{2}\Big(\int\frac{du}{u}-\int\frac{du}{u+1}\Big)$$ $$I=\frac{1}{2}\Big(\ln|u|-\ln|u+1|\Big)+C$$ $$I=\frac{1}{2}[\ln(x^2)-\ln(x^2+1)]+C$$ (since $x^2\ge0$) $$I=\frac{1}{2}\ln\frac{x^2}{x^2+1}+C$$
Therefore, $$A=\frac{1}{2}\ln\frac{x^2}{x^2+1}-\frac{\tan^{-1}x}{x}+C$$
