For starters, substitute ${x}^{2}\mapsto x,\text{}$ as follows :

$\begin{array}{rl}I\text{}& =\text{}\phantom{\rule{1em}{0ex}}{\int}_{0}^{\mathrm{\infty}}\text{}x\text{}\mathrm{ln}(1+{x}^{2})\text{}{e}^{-Bx}\text{}dx\text{}=\text{}\frac{1}{2}\phantom{\rule{1em}{0ex}}{\int}_{0}^{\mathrm{\infty}}\mathrm{ln}(1\text{}+\text{}{x}^{2})\text{}{e}^{-B\sqrt{{x}^{2}}}\text{}d({x}^{2})\text{}=\text{}\\ \\ \text{}& =\text{}\frac{1}{2}\text{}{\int}_{0}^{\mathrm{\infty}}\phantom{\rule{1em}{0ex}}\mathrm{ln}(1+x)\text{}{e}^{-B\sqrt{x}}\text{}dx\text{}=\text{}\frac{1}{2}\text{}{\textstyle [}{\int}_{0}^{\mathrm{\infty}}\mathrm{ln}(1+Ax)\text{}{e}^{-B\sqrt{x}}\text{}dx{{\textstyle ]}}_{A=1}\text{}=\text{}\frac{J(1)}{2}\end{array}$

Now, employ differentiation under the integral sign :

$\begin{array}{rl}{J}^{\prime}(A)\text{}& =\text{}{\int}_{0}^{\mathrm{\infty}}\frac{x}{1+Ax}\text{}{e}^{-B\sqrt{x}}\text{}dx\text{}=\text{}\frac{{d}^{2}}{d{B}^{2}}\text{}{\int}_{0}^{\mathrm{\infty}}\frac{1}{1+Ax}\text{}{e}^{-B\sqrt{x}}\text{}dx\text{}=\text{}\\ \\ \text{}& =\text{}\frac{{d}^{2}}{d{B}^{2}}\text{}{\textstyle \{}\frac{1}{A}\text{}{\textstyle [}\mathrm{sin}\frac{B}{\sqrt{A}}{\textstyle (}\pi -2\text{Si}\frac{B}{\sqrt{A}}{\textstyle )}-2\text{}\mathrm{cos}\frac{B}{\sqrt{A}}\text{Ci}{\displaystyle \frac{B}{\sqrt{A}}}{\textstyle ]}{\textstyle \}},\end{array}$

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

$\frac{J(A)}{2}\text{}=\text{}\frac{{d}^{2}}{d{B}^{2}}{\textstyle [}{\text{Ci}}^{2}\text{}{\displaystyle \frac{B}{\sqrt{A}}}+{\text{Si}}^{2}\text{}{\displaystyle \frac{B}{\sqrt{A}}}-\pi \text{Si}{\displaystyle \frac{B}{\sqrt{A}}}{\textstyle ]},$

which, when evaluated at A=1, gives $\text{}I\text{}=\text{}{\displaystyle \frac{J(1)}{2}}\text{}=\text{}{\displaystyle \frac{{d}^{2}}{d{B}^{2}}}\text{}{\textstyle [}{\text{Ci}}^{2}\text{}B+{\text{Si}}^{2}\text{}B-\pi \text{Si}B{\textstyle ]},$

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