Primitive of Reciprocal of 1 plus x squared/Arctangent Form
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Theorem
- $\ds \int \frac {\d x} {1 + x^2} = \arctan x + C$
where $C$ is an arbitrary constant.
Proof 1
From Primitive of $\dfrac 1 {x^2 + a^2}$: Arctangent Form:
- $\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$
The result follows by setting $a = 1$.
$\blacksquare$
Proof 2
\(\ds \map {\dfrac \d {\d x} } {\arctan x}\) | \(=\) | \(\ds \dfrac 1 {1 + x^2}\) | Derivative of Arctangent Function | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \dfrac {\d x} {1 + x^2}\) | \(=\) | \(\ds \arctan x + C\) | Definition of Primitive (Calculus) |
$\blacksquare$