(18)int x^2arctan xdx.

设 $u = \arctan x$，$dv = x^2 \, dx$，则 $du = \frac{1}{1+x^2} \, dx$，$v = \frac{x^3}{3}$。分部积分得： \[ \int x^2 \arctan x \, dx = \frac{x^3}{3} \arctan x - \frac{1}{3} \int \frac{x^3}{1+x^2} \, dx \] 将被积函数拆分： \[ \frac{x^3}{1+x^2} = x - \frac{x}{1+x^2} \] 则： \[ \int \frac{x^3}{1+x^2} \, dx = \int x \, dx - \int \frac{x}{1+x^2} \, dx = \frac{x^2}{2} - \frac{1}{2} \ln(1+x^2) \] 代回原式： \[ \int x^2 \arctan x \, dx = \frac{x^3}{3} \arctan x - \frac{1}{3} \left( \frac{x^2}{2} - \frac{1}{2} \ln(1+x^2) \right) + C \] 化简得： \[ \boxed{\frac{1}{3} x^3 \arctan x - \frac{1}{6} x^2 + \frac{1}{6} \ln(1+x^2) + C} \]