Primitive of x by Arctangent of x/Proof 1
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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 $a = 1$.
$\blacksquare$