Compare parts with an arctangent substitution algebraically

Parts with $u=\tan^{-1}x$, $dv=x\,dx$ gives $\frac{x^2}2\tan^{-1}x-\frac12\int\dfrac{x^2}{1+x^2}dx$. The source callout uses long division: $x^2/(1+x^2)=1-1/(1+x^2)$. Hence the answer is $\frac{x^2}2\tan^{-1}x-\frac x2+\frac12\tan^{-1}x+C_1$.

$\dfrac d{d\theta}(\tan^2\theta)=2\tan\theta\sec^2\theta$.

With $\theta=\tan^{-1}x$, $x=\tan\theta$ and $dx=\sec^2\theta\,d\theta$. The integral becomes $\int\theta\tan\theta\sec^2\theta\,d\theta$. Use the preceding derivative in parts to get $\frac12\theta\tan^2\theta-\frac12\int\tan^2\theta\,d\theta$. Since $\tan^2\theta=\sec^2\theta-1$, this is $\frac12\theta\tan^2\theta-\frac12(\tan\theta-\theta)+C_2$. Substitute back to obtain $\frac{x^2}2\tan^{-1}x-\frac x2+\frac12\tan^{-1}x+C_2$.

Subtract the two displayed results. Every x-dependent term cancels, leaving $C_1-C_2$, a constant, as required. The printed methods give the same explicit nonconstant expression.