Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Proof 4

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