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

Question

Aicharya · Accepted Answer

$\begin{aligned} & I=\int_{-1}^2 \log _e\left(x+\sqrt{x^2+1}ight) d x \ & =x \log _e\left(x+\sqrt{x^2+1}ight)-\int_{-1}^2\left(\frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}ight) d x \ & =x \log _e\left(x+\sqrt{x^2+1}ight)-\int_{-1}^2 \frac{x}{\sqrt{x^2+1}} d x\end{aligned}$ $\begin{aligned} & =x \log _e\left(x+\sqrt{x^2+1}ight)-\left.\sqrt{x^2+1}ight|_{-1} ^2 \ & =\left(2 \log _e(2+\sqrt{5})-\sqrt{5}ight) \ & -\left(-\log _e(-1+\sqrt{2})-\sqrt{2}ight) \ & =\log _e(2+\sqrt{5})^2-\sqrt{5}+\log _e(\sqrt{2}-1)+\sqrt{2} \ & =\log _e(2+\sqrt{5})^2-\sqrt{5}+\log _e(\sqrt{2}-1)+\sqrt{2}\end{aligned}$ $\begin{aligned} & =\sqrt{2}-\sqrt{5}+\log _e\left(\frac{(2+\sqrt{5})^2}{\sqrt{2+1}}ight) \ & =\sqrt{2}-\sqrt{5}+\log _e\left(\frac{9+4 \sqrt{5}}{\sqrt{2+1}}ight)\end{aligned}$

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

Solution