# Primitive of Reciprocal of x squared plus a squared/Arctangent Form

## Theorem

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

where $a$ is a non-zero constant.

## Proof 1

Let:

$a \tan \theta = x$

for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$.

From Shape of Tangent Function, this substitution is valid for all real $x$.

Then:

 $\ds x$ $=$ $\ds a \tan \theta$ from above $\ds \leadsto \ \$ $\ds \frac {\d x} {\d \theta}$ $=$ $\ds a \sec^2 \theta$ Derivative of Tangent Function $\ds \leadsto \ \$ $\ds \int \frac 1 {x^2 + a^2} \rd x$ $=$ $\ds \int \frac {a \ \sec^2 \theta} {a^2 \tan^2 \theta + a^2} \rd \theta$ Integration by Substitution $\ds$ $=$ $\ds \frac a {a^2} \int \frac {\sec^2 \theta} {\tan^2 \theta + 1} rd \theta$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds \frac 1 a \int \frac {\sec^2 \theta} {\sec^2 \theta} \rd \theta$ Difference of Squares of Secant and Tangent $\ds$ $=$ $\ds \frac 1 a \int \rd \theta$ $\ds$ $=$ $\ds \frac 1 a \theta + C$ Integral of Constant

As $\theta$ was stipulated to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:

$\tan \theta = \dfrac x a \iff \theta = \arctan \dfrac x a$

Thus:

$\ds \int \frac 1 {x^2 + a^2} \rd x = \frac 1 a \arctan \frac x a + C$

$\Box$

When $a = 0$, both $\dfrac x a$ and $\dfrac 1 a$ are undefined.

However, consider the limit of the above primitive as $a \to 0$:

 $\ds \lim_{a \mathop \to 0} \frac 1 a \arctan {\frac x a}$ $=$ $\ds \lim_{a \mathop \to 0} \frac {\arctan {\frac x a} } a$ $\ds \leadsto \ \$ $\ds$ $=$ $\ds \lim_{a \mathop \to 0} \frac {-x a^{-2} } {1 + \frac {x^2} {a^2} }$ L'Hôpital's Rule, Derivative of Arctangent Function $\ds$ $=$ $\ds \lim_{a \mathop \to 0} \frac {a^{-2} } {a^{-2} } \frac {-x} {x^2 + a^2}$ $\ds$ $=$ $\ds -\frac 1 x$

This corresponds with the result:

$\ds \int \frac 1 {x^2} \rd x = \frac {-1} x + C$

which follows from Primitive of Power.

$\blacksquare$

## Proof 2

We have that $x^2 + a^2$ is in the form $a x^2 + b x + c$, where $b^2 - 4 a c < 0$.

Thus from Primitive of $\dfrac 1 {a x^2 + b x + c}$ for $b^2 - 4 a c > 0$:

$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$

setting $a := 1, b := 0, c := a^2$:

 $\ds \int \frac 1 {x^2 + a^2} \rd x$ $=$ $\ds \dfrac 2 {\sqrt {4 a^2 - 0} } \map \arctan {\dfrac {2 x + 0} {\sqrt {4 a^2} } } + C$ Primitive of $\dfrac 1 {a x^2 + b x + c}$ $\ds$ $=$ $\ds \frac 1 a \arctan {\frac x a} + C$ simplifying

$\blacksquare$

## Proof 3

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

$\blacksquare$