8. (2022·江苏)求不定积分int xarctan(1)/(x)dx.

令 $u = \arctan \frac{1}{x}$，则 $du = -\frac{1}{x^2 + 1}dx$。取 $dv = xdx$，则 $v = \frac{x^2}{2}$。应用分部积分法：
$\int x \arctan \frac{1}{x} \, dx = \frac{x^2}{2} \arctan \frac{1}{x} - \int \frac{x^2}{2} \left( -\frac{1}{x^2 + 1} \right) \, dx = \frac{x^2}{2} \arctan \frac{1}{x} + \frac{1}{2} \int \frac{x^2}{x^2 + 1} \, dx.$
而 $\frac{x^2}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}$，故：
$\int \frac{x^2}{x^2 + 1} \, dx = x - \arctan x.$
代入得：
$\int x \arctan \frac{1}{x} \, dx = \frac{x^2}{2} \arctan \frac{1}{x} + \frac{1}{2} (x - \arctan x) + C = \frac{x^2}{2} \arctan \frac{1}{x} + \frac{x}{2} - \frac{1}{2} \arctan x + C.$
答案：
$\boxed{\frac{x^2}{2} \arctan \frac{1}{x} + \frac{x}{2} - \frac{1}{2} \arctan x + C}$