Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Proof 4
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Theorem
- $\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$
Proof
\(\ds \map {\dfrac \d {\d x} } {\frac 1 a \arctan \frac x a}\) | \(=\) | \(\ds \frac 1 a \paren {\dfrac a {a^2 + x^2} }\) | Derivative of Arctangent Function: Corollary and Derivative of Constant Multiple | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {a^2 + x^2}\) | simplification | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\d x} {x^2 + a^2}\) | \(=\) | \(\ds \frac 1 a \arctan \frac x a + C\) | Fundamental Theorem of Calculus |
$\blacksquare$
Sources
- 1945: A. Geary, H.V. Lowry and H.A. Hayden: Advanced Mathematics for Technical Students, Part I ... (previous) ... (next): Chapter $\text {III}$: Integration: Integration