Primitive of Reciprocal of 1 plus x squared/Arctangent Form

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)

$\blacksquare$