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I can not calculate this integration please help me...

Note that this integral is part of complex dual integration and i must obtain that with constants. I tried by MATLAB but ...

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$$ \int_0^\infty\ln\left(\sqrt{x} + a\right)b\mathrm{e}^{-bx}dx = -\int_0^\infty \ln\left(\sqrt{x} + a\right)\frac{d}{dx}\mathrm{e}^{-bx}dx $$ integrating by parts we find $$ -\left[\mathrm{e}^{-bx}\ln\left(\sqrt{x} + a\right)\right]_0^\infty + \int_0^\infty \frac{1}{\sqrt{x}+a}\frac{1}{\sqrt{x}}\mathrm{e}^{-bx}dx $$ we have $$ \frac{1}{\sqrt{x}+a}\frac{1}{\sqrt{x}} = \frac{1}{a}\left[\frac{1}{\sqrt{x} + a} -\frac{1}{\sqrt{x}}\right] $$ so ultimately you are trying to solve $$ \int_0^\infty\frac{1}{\sqrt{x} + a}\mathrm{e}^{-bx}dx $$ and $$ \int_0^\infty\frac{1}{\sqrt{x}}\mathrm{e}^{-bx}dx $$ This $$ \frac{1}{\sqrt{b}}\int_0^\infty\frac{1}{\sqrt{u}}\mathrm{e}^{-u}du = \frac{1}{\sqrt{b}}\Gamma(1/2) = \sqrt{\frac{\pi}{b}} $$ The other integral is not great..

To finish off $$ \left[\mathrm{e}^{-bx}\ln\left(\sqrt{x} + a\right)\right]_0^\infty = -\ln(a) $$ Since $$ \lim_{x\to\infty}\frac{\ln x}{x} \to 0 $$ and $$ \lim_{x\to\infty}\frac{\mathrm{e}^x}{x}\to \infty $$ then we can conclude that the term in the bracket tends to $0$.

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    and where is the $\int_{0}^{\infty} \frac{1}{\sqrt(x) + a} e^{-b x} dx$ ? – 2017-01-10
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    @MrBlue that is the hard part :-S – 2017-01-10
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    My mean is Why it is not great? the integral is part of another integral and eliminating that can be trouble ... – 2017-01-10
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    @tired Sadly I can not solve this one! MrBlue, I was trying to show where the whole integral probably breaks down. – 2017-01-10
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    The remaining integral in question reads, after a simple sub, $$ 2\int_0^{\infty}\frac{x}{x+a}e^{-bx^2}=2\int_0^{\infty}e^{-bx^2}-2a\int_0^{\infty}\frac{1}{x+a}e^{-bx^2} $$ A calculation of the last integral might be extracted from here: http://math.stackexchange.com/questions/1412761/another-integral-related-to-fresnel-integrals/1417077#1417077 – 2017-01-10
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    @tired If you want to provide a completed answer then go ahead! I did not want to proceed further given the OP said it broke down with Matlab! But alas all excuses, and I enjoyed reading your answer! – 2017-01-10
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    @tired Unfortunately I don't understand this http://math.stackexchange.com/questions/1412761/another-integral-related-to-fresnel-integrals/1417077#1417077 – 2017-01-10
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    @MrBlue where exactly you have difficulties? – 2017-01-11
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    @tired It was Goodwin-Staton integral. I find that's solution. Thank you – 2017-01-12