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


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

$\blacksquare$