Answer
$$\int x\tan^{-1}xdx=\frac{(x^2+1)}{2}\tan^{-1}x-\frac{1}{2}x+C$$
Work Step by Step
$$A=\int x\tan^{-1}xdx$$
Use Formula 108, which states that $$\int x^n\tan^{-1}axdx=\frac{x^{n+1}}{n+1}\tan^{-1}ax-\frac{a}{n+1}\int\frac{x^{n+1}dx}{1+a^2x^2}$$
for $n=1$ and $a=1$ here:
$$A=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\int\frac{x^2dx}{1+x^2}$$ $$A=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\Big(\int\frac{1+x^2}{1+x^2}dx-\int\frac{1dx}{1+x^2}\Big)$$ $$A=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\Big(\int dx-\int\frac{dx}{1+x^2}\Big)$$ $$A=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\Big(x-\int\frac{dx}{1+x^2}\Big)$$
According to Formula 32, which states that
$$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C$$
for $a=1$:
$$A=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\Big(x-\tan^{-1}x\Big)+C$$ $$A=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}x+\frac{1}{2}\tan^{-1}x+C$$ $$A=\frac{(x^2+1)}{2}\tan^{-1}x-\frac{1}{2}x+C$$
