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

Theorem

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

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 $x^m \ln \left({x^2 - a^2}\right)$