Primitive of Reciprocal of 1 plus x squared/Arctangent Form/Proof 2
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Theorem
- $\ds \int \frac {\d x} {1 + x^2} = \arctan x + C$
Proof
| \(\ds \map {\dfrac \d {\d x} } {\arctan x}\) | \(=\) | \(\ds \dfrac 1 {1 + x^2}\) | Derivative of Arctangent Function | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \int \dfrac {\d x} {1 + x^2}\) | \(=\) | \(\ds \arctan x + C\) | Definition of Primitive (Calculus) |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Integration
- 1967: Michael Spivak: Calculus ... (previous) ... (next): Part $\text {III}$: Derivatives and Integrals: Chapter $18$: Integration in Elementary Terms