Mathematics - Definite Integration Question with Solution | TestHub

$\begin{array}{rl}& I={\int }_{-1}^2{\log }_e\left(x+\sqrt{x^2+1}\right)dx\\ & =x{\log }_e\left(x+\sqrt{x^2+1}\right)-{\int }_{-1}^2\left(\frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}\right)dx\\ & =x{\log }_e\left(x+\sqrt{x^2+1}\right)-{\int }_{-1}^2\frac{x}{\sqrt{x^2+1}}dx\end{array}$ $\begin{array}{rl}& =x{\log }_e\left(x+\sqrt{x^2+1}\right)-{\sqrt{x^2+1}|}_{-1}^2\\ & =\left(2{\log }_e(2+\sqrt{5})-\sqrt{5}\right)\\ & -\left(-{\log }_e(-1+\sqrt{2})-\sqrt{2}\right)\\ & ={\log }_e(2+\sqrt{5}{)}^2-\sqrt{5}+{\log }_e(\sqrt{2}-1)+\sqrt{2}\\ & ={\log }_e(2+\sqrt{5}{)}^2-\sqrt{5}+{\log }_e(\sqrt{2}-1)+\sqrt{2}\end{array}$ $\begin{array}{rl}& =\sqrt{2}-\sqrt{5}+{\log }_e\left(\frac{(2+\sqrt{5}{)}^2}{\sqrt{2+1}}\right)\\ & =\sqrt{2}-\sqrt{5}+{\log }_e\left(\frac{9+4\sqrt{5}}{\sqrt{2+1}}\right)\end{array}$