Primitive of Reciprocal of 1 plus x squared/Arctangent Form/Proof 1
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Corollary to Primitive of $\frac 1 {x^2 + a^2}$: Arctangent Form
- $\ds \int \frac {\d x} {1 + x^2} = \arctan x + C$
Proof
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 $a = 1$.
$\blacksquare$