The value of the integral _-1^2 _e(x+x^2+1)dx is

Question

The value of the integral $\int _{-1}^{2}{\log }_{e}(x+\sqrt{{x}^{2}+1})dx$ is

Accepted Answer

$\begin{gathered} \int _{-1}^{2}{\log }_{e}(x+\sqrt{{x}^{2}+1})dx \ ={[x{\log }_{e}(x+\sqrt{{x}^{2}+1})]}_{2}^{-1}-\int _{-1}^{2}\frac{x}{(x+\sqrt{{x}^{2}+1})}(1+\frac{x}{\sqrt{{x}^{2}+1}})dx \ =2{\log }_{2}(2+\sqrt{5})+{\log }_{e}(\sqrt{2}-1)-\int _{-1}^{2}\frac{x}{\sqrt{{x}^{2}+1}}dx \ ={\log }_{e}[(2+\sqrt{5}{)}^{2}(\sqrt{2}-1)]-{[\sqrt{{x}^{2}+1}]}_{2}^{-1} \ ={\log }_{e}[\frac{9+4\sqrt{5}}{1+\sqrt{2}}]-\sqrt{5}+\sqrt{2} \ =\sqrt{2}-\sqrt{5}+{\log }_{e}[\frac{9+4\sqrt{5}}{1+\sqrt{2}}] \end{gathered}$

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