Primitive of Power of x by Logarithm of x squared plus a squared

Theorem

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

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:

and let:

Then:

Also see

 * Primitive of $x^m \map \ln {x^2 - a^2}$