12.1. a) Calculate:

$$
\int_{1}^{2}\left(\frac{1}{x}-\arcsin \frac{1}{x}\right) d x
$$

b) Compare the numbers: $\frac{\pi}{6} \text{ and } \ln \frac{2+\sqrt{3}}{2}$.
## Solution:

a)

$$
\begin{aligned}
& I=\int_{1}^{2}\left(\frac{1}{x}-\arcsin \frac{1}{x}\right) d x=\left.\ln x\right|_{1} ^{2}-\left.x \arcsin \frac{1}{x}\right|_{1} ^{2}+\int_{1}^{2} x \cdot \frac{1}{\sqrt{1-\frac{1}{x^{2}}}} \cdot\left(-\frac{1}{x^{2}}\right) d x= \\
& \ln 2+\frac{\pi}{6}-\int_{1}^{2} \frac{1}{\sqrt{x^{2}-1}} d x=\ln 2+\frac{\pi}{6}-\left.\ln \left(x+\sqrt{x^{2}-1}\right)\right|_{1} ^{2}=\frac{\pi}{6}-\ln \frac{2+\sqrt{3}}{2}
\end{aligned}
$$

For $x \in [1; 2]$, $\arcsin \frac{1}{x} > \frac{1}{x}$, which implies $I=\int_{1}^{2}\left(\frac{1}{x}-\arcsin \frac{1}{x}\right) d x < 0$. Using the value of the integral $I$ obtained in part a), we get

$$
\frac{\pi}{6} < \ln \frac{2+\sqrt{3}}{2}
$$