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Question Number 128079 by liberty last updated on 04/Jan/21

$$\mathrm{\Omega }=\int \ln (\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2})\mathrm{dx}$$

Answered by bemath last updated on 04/Jan/21

$$\mathrm{\Omega }=\mathrm{x}\ln (\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2})-\int \frac{\mathrm{x}(1+\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}})}{\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}\mathrm{dx}$$$$\mathrm{\Omega }=\mathrm{x}\ln (\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2})+\frac{1}{{\mathrm{a}}^2}\int \mathrm{x}(\mathrm{x}-\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2})\left(\frac{\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}\right)\mathrm{dx}$$$$\mathrm{\Omega }=\mathrm{x}\ln (\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2})+\frac{1}{{\mathrm{a}}^2}\int \frac{\mathrm{x}(-{\mathrm{a}}^2)}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}\mathrm{dx}$$$$\mathrm{\Omega }=\mathrm{x}\ln (\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2})-\frac{1}{2}\int \frac{\mathrm{d}({\mathrm{x}}^2+{\mathrm{a}}^2)}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}$$$$\mathrm{\Omega }=\mathrm{x}\ln (\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2})-\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}+\mathrm{C}$$$$$$

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