Answer
$\displaystyle \frac{d}{dx}$(RHS) = integrand on the LHS,
so the statement is valid. See below.
Work Step by Step
Differentiate the RHS and see if it equals $\displaystyle \frac{\tan^{-1}x}{x}$
$\displaystyle \frac{d}{dx}[\ln x]=\frac{1}{x}$
$\displaystyle \frac{d}{dx}[\ln(1+x^{2})]=$ (use chain rule) $=\displaystyle \frac{1}{1+x^{2}}\cdot 2x$
$\displaystyle \frac{d}{dx}[\frac{\tan^{-1}x}{x}]$ = (use quotient rule, table 7-3) $=\displaystyle \frac{\frac{1}{1+x^{2}}\cdot x-(1)\cdot\tan^{-1}x}{x^{2}}$
$=\displaystyle \frac{x-(1+x^{2})\tan^{-1}x}{x^{2}(1+x^{2})}$
$RHS=\displaystyle \frac{1}{x}-\frac{1}{2}\cdot\frac{2x}{1+x^{2}}-\frac{x-(1+x^{2})\tan^{-1}x}{x^{2}(1+x^{2})}$
$=\displaystyle \frac{x(1+x^{2}) -x\cdot x^{2}-(x-(1+x^{2})\tan^{-1}x)}{x^{2}(1+x^{2})}$
$=\displaystyle \frac{x+x^{3} -x^{3}-x+(1+x^{2})\tan^{-1}x}{x^{2}(1+x^{2})}$
$=\displaystyle \frac{(1+x^{2})\tan^{-1}x}{x^{2}(1+x^{2})}$
$=\displaystyle \frac{\tan^{-1}x}{x}$
Since $\displaystyle \frac{d}{dx}$(RHS) = integrand on the LHS, the statement is valid.
