# Primitive of Arccotangent of x over a over x squared

## Theorem

$\displaystyle \int \frac {\arccot \frac x a \rd x} {x^2} = \frac {-\arccot \frac x a} x + \frac 1 {2 a} \map \ln {\frac {x^2 + a^2} {x^2} } + C$

## Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

 $\displaystyle u$ $=$ $\displaystyle \arccot \frac x a$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d u} {\d x}$ $=$ $\displaystyle \frac {-a} {x^2 + a^2}$ Derivative of $\arccot \dfrac x a$

and let:

 $\displaystyle \frac {\d v} {\d x}$ $=$ $\displaystyle \frac 1 {x^2}$ $\displaystyle \leadsto \ \$ $\displaystyle v$ $=$ $\displaystyle \frac {-1} x$ Primitive of Power

Then:

 $\displaystyle \int \frac {\arccot \frac x a \rd x} {x^2}$ $=$ $\displaystyle \arccot \frac x a \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac {-a} {x^2 + a^2} } \rd x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac {-\arccot \frac x a} x - a \int \frac {\d x} {x \paren {x^2 + a^2} } \rd x + C$ simplifying $\displaystyle$ $=$ $\displaystyle \frac {-\arccot \frac x a} x - \frac 1 a \paren {\frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } } + C$ Primitive of $\dfrac 1 {x \paren {x^2 + a^2} }$ $\displaystyle$ $=$ $\displaystyle \frac {-\arccot \frac x a} x + \frac 1 {2 a} \map \ln {\frac {x^2 + a^2} {x^2} } + C$ Logarithm of Reciprocal

$\blacksquare$