Find an approximate equivalent. int_0^(+oo) x log(1+x^2)e^(−Bx) dx

This integral sounds quite complex and I could not find an approximate equivalent.
${\int }_{0}^{+\mathrm{\infty }}x\mathrm{log}\left(1+{x}^{2}\right)\phantom{\rule{thinmathspace}{0ex}}{e}^{-Bx}\phantom{\rule{thinmathspace}{0ex}}dx$
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jennasyliang4tr
For starters, substitute as follows :

Now, employ differentiation under the integral sign :

which, when integrated with regard to A, yields the rather beautiful expression :

which, when evaluated at A=1, gives