Evaluating ln and exponential integral

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$$I = \int_{0}^\infty \dfrac{e^{-\sqrt{Ax}} \ln \left(1+x\right)}{\sqrt{x}} dx , \quad A>0.$$

Any hint for solving this integral, I think Gauss quadrature cannot be applied to this. Matlab is giving numerical answers to this, that means its solvable. Any help would be highly appreciated.

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Your question is equivalent to finding $$ g(B) = \int_{0}^{+\infty} \log(1+x^2)\,e^{- B x}\,dx $$ for $B>0$, i.e. to finding the Laplace transform of $\log(1+x^2)$. By integration by parts we have $$ g(B) = \frac{2}{B}\int_{0}^{+\infty}\frac{x}{1+x^2}\,e^{-B x}\,dx $$ and by partial fraction decomposition that equals $$ \frac{2}{B}\left[\frac{\pi}{2}\sin(B)-\sin(B)\text{Si}(B)-\cos(B)\text{Ci}(B)\right] $$ where $\text{Si}$ and $\text{Ci}$ are the sine/cosine integrals. A numerical evaluation is indeed simpler from the previous integral representation, since both $\frac{x}{1+x^2}$ and $e^{-Bx}$ are smooth and decreasing functions. For instance, by the Cauchy-Schwarz inequality

$$ g(B)\leq\frac{2}{B}\sqrt{\int_{0}^{+\infty}\frac{x\,dx}{(1+x^2)^2}\int_{0}^{+\infty}x\,e^{-2B x}\,dx}=\frac{1}{B^2\sqrt{2}}. $$