问题及解答

求定积分

\[\int_1^2\frac{\sqrt{x^2-1}}{x^2}dx.\]

\[
\begin{split}
\int_1^2\frac{\sqrt{x^2-1}}{x^2}dx&=-\int_1^2\sqrt{x^2-1}d\frac{1}{x}\\
&=-\biggl[\frac{\sqrt{x^2-1}}{x}\biggr|_{1}^{2}-\int_1^2\frac{1}{x}d\sqrt{x^2-1}\biggr]\\
&=\int_1^2\frac{1}{x}\cdot\frac{2x}{2\sqrt{x^2-1}}dx-\frac{\sqrt{3}}{2}\\
&=\int_1^2\frac{1}{\sqrt{x^2-1}}dx-\frac{\sqrt{3}}{2}.
\end{split}
\]

 其中

\[
\begin{split}
\int_1^2\frac{1}{\sqrt{x^2-1}}dx&\stackrel{x=\sec t}{=}\int_0^{\frac{\pi}{3}}\frac{1}{\tan t}d\sec t=\int_0^{\frac{\pi}{3}}\frac{\sec t\cdot\tan t}{\tan t}dt=\int_0^{\frac{\pi}{3}}\sec t dt\\
&=\ln |\sec t+\tan t|\biggl|_{0}^{\frac{\pi}{3}}\\
&=\ln(2+\sqrt{3}).
\end{split}
\]

这里倒数第二个等号请查看问题1410.

因此, 原积分的值为 $\ln(2+\sqrt{3})-\frac{\sqrt{3}}{2}$.