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

## 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 $t \to \frac{x}{a}$ | $\quad$ |

$\blacksquare$