Integrals: Integration by Parts Inverse Tangent

Let $u = \arctan x, \, dv = x \, dx \implies du = \frac{1}{1 + x^2} dx, \, v = \frac{x^2}{2}$: $\int x \arctan x \, dx = \frac{x^2}{2} \arctan x - \frac{1}{2} \int \frac{x^2}{1 + x^2} \, dx$

$\frac{x^2}{1 + x^2} = \frac{(x^2 + 1) - 1}{1 + x^2} = 1 - \frac{1}{1 + x^2}$ $\int \frac{x^2}{1 + x^2} \, dx = \int 1 \, dx - \int \frac{1}{1 + x^2} \, dx = x - \arctan x$

$= \frac{x^2}{2} \arctan x - \frac{1}{2}(x - \arctan x) + C = \frac{x^2 + 1}{2} \arctan x - \frac{x}{2} + C$