Primitive of Arctangent Function
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Theorem
- $\ds \int \arctan x \rd x = x \arctan x - \frac {\map \ln {x^2 + 1} } 2 + C$
Proof
From Primitive of $\arctan \dfrac x a$:
- $\ds \int \arctan \frac x a \rd x = x \arctan \frac x a - \frac a 2 \map \ln {x^2 + a^2} + C$
The result follows by setting $a = 1$.
$\blacksquare$
Also presented as
This result can also be presented as:
- $\ds \int \arctan x \rd x = x \arctan x - \ln \sqrt {x^2 + 1} + C$
Also see
Sources
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): arc-tangent
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Appendix: Table $2$: Integrals
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Appendix: Table $2$: Integrals