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

Jump to navigation Jump to search

## 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 $t \to \dfrac x a$

$\blacksquare$