JEE MAIN - Mathematics (2024 - 9th April Evening Shift - No. 11)

The value of the integral $$\int_\limits{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x$$ is
$$\sqrt{5}-\sqrt{2}+\log _e\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)$$
$$\sqrt{2}-\sqrt{5}+\log _e\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)$$
$$\sqrt{5}-\sqrt{2}+\log _e\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$$
$$\sqrt{2}-\sqrt{5}+\log _e\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$$

Explanation

$$\begin{aligned} & \int_\limits{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x \\ & =\left[x \log _e\left(x+\sqrt{x^2+1}\right)\right]_{-1}^2-\int_\limits{-1}^2 \frac{x}{\left(x+\sqrt{x^2+1}\right)}\left(1+\frac{x}{\sqrt{x^2+1}}\right) d x \\ & =2 \log _2(2+\sqrt{5})+\log _e(\sqrt{2}-1)-\int_\limits{-1}^2 \frac{x}{\sqrt{x^2+1}} d x \\ & =\log _e\left[(2+\sqrt{5})^2(\sqrt{2}-1)\right]-\left[\sqrt{x^2+1}\right]_{-1}^2 \\ & =\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right]-\sqrt{5}+\sqrt{2} \\ & =\sqrt{2}-\sqrt{5}+\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right] \end{aligned}$$

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