The value of the integral $\int_\limits{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x$ is
Solution
<p>$$\begin{aligned}
& \int_\limits{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x \\
& =\left[x \log _e\left(x+\sqrt{x^2+1}\right)\right]_{-1}^2-\int_\limits{-1}^2 \frac{x}{\left(x+\sqrt{x^2+1}\right)}\left(1+\frac{x}{\sqrt{x^2+1}}\right) d x \\
& =2 \log _2(2+\sqrt{5})+\log _e(\sqrt{2}-1)-\int_\limits{-1}^2 \frac{x}{\sqrt{x^2+1}} d x \\
& =\log _e\left[(2+\sqrt{5})^2(\sqrt{2}-1)\right]-\left[\sqrt{x^2+1}\right]_{-1}^2 \\
& =\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right]-\sqrt{5}+\sqrt{2} \\
& =\sqrt{2}-\sqrt{5}+\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right]
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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