Primitive of Arctangent of x over a over x squared

Theorem

 * $\displaystyle \int \frac {\arctan \frac x a \ \mathrm d x} {x^2} = \frac {-\arctan \frac x a} x - \frac 1 {2 a} \ln \left({\frac {x^2 + a^2} {x^2} }\right) + C$

Proof
With a view to expressing the primitive in the form:
 * $\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

and let:

Then:

Also see

 * Primitive of $\dfrac {\arcsin \dfrac x a} {x^2}$


 * Primitive of $\dfrac {\arccos \dfrac x a} {x^2}$


 * Primitive of $\dfrac {\operatorname{arccot} \dfrac x a} {x^2}$


 * Primitive of $\dfrac {\operatorname{arcsec} \dfrac x a} {x^2}$


 * Primitive of $\dfrac {\operatorname{arccsc} \dfrac x a} {x^2}$