Primitive of Arccotangent of x over a/Proof 2

Theorem

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

With a view to expressing the primitive in the form:

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

let:

$\ds u$

$=$

$\ds \arccot \frac x a$

$\ds \leadsto \ \ $

$\ds \frac {\d u} {\d x}$

$=$

$\ds \frac {-a} {x^2 + a^2}$

and let:

$\ds \frac {\d v} {\d x}$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds v$

$=$

$\ds x$

Then:

$\ds \int \arccot \frac x a \rd x$

$=$

$\ds x \arccot \frac x a - \int x \paren {\frac {-a} {x^2 + a^2} } \rd x + C$

$\ds $

$=$

$\ds x \arccot \frac x a + a \int \frac {x \rd x} {x^2 + a^2} + C$

$\ds $

$=$

$\ds x \arccot \frac x a + a \paren {\frac 1 2 \map \ln {x^2 + a^2} } + C$

$\ds $

$=$

$\ds x \arccot \frac x a + \frac a 2 \map \ln {x^2 + a^2} + C$

simplifying

$\blacksquare$