Primitive of Reciprocal of x squared plus a squared/Proof 3

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Theorem

$\displaystyle \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$


Proof

\(\displaystyle \int \frac {\d x} {x^2 + a^2}\) \(=\) \(\displaystyle \frac 1 a \int \frac {\d t} {t^2 + 1}\) Substitution of $x \to a t$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \int \frac {\d t} {\paren {1 + i t} \paren {1 - i t} }\) Factoring
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 a} \paren {\int \frac {\d t} {1 + i t} + \int \frac {\d t} {1 - i t} }\) Definition of Partial Fractions Expansion
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 a} \paren {i \, \map \ln {1 - i t} - i \, \map \ln {1 + i t} } + C\) Primitive of Reciprocal
\(\displaystyle \) \(=\) \(\displaystyle \frac i {2 a} \map \ln {\frac {1 - i t} {1 + i t} } + C\) Sum of Logarithms
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \arctan \frac x a + C\) Arctangent Logarithmic Formulation and substituting back $t \to \dfrac x a$

$\blacksquare$