Primitive of x by Arctangent of x/Proof 1

Theorem

$\ds \int x \arctan x \rd x = \frac {x^2 + 1} 2 \arctan x - \frac x 2 + C$

Proof

From Primitive of $x \arctan \dfrac x a$:

$\ds \int x \arctan \frac x a \rd x = \frac {x^2 + a^2} 2 \arctan \frac x a - \frac {a x} 2 + C$

The result follows on setting $a = 1$.

$\blacksquare$