## Problem Statement

Calculate the definite integral:

$$
\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}}
$$
## Solution

$$
\begin{aligned}
& \int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}}=\int_{\sqrt{2}}^{2} \frac{d x}{x^{2} \cdot \frac{1}{x} \cdot \sqrt{x^{2}-1}}=\int_{\sqrt{2}}^{2} \frac{d x}{x^{2} \cdot \sqrt{1-\frac{1}{x^{2}}}}=-\int_{\sqrt{2}}^{2} \frac{d\left(\frac{1}{x}\right)}{\sqrt{1-\frac{1}{x^{2}}}}= \\
= & \int_{\sqrt{2}}^{2} \frac{d x}{x^{2} \cdot \sqrt{1-\frac{1}{x^{2}}}}=-\int_{\sqrt{2}}^{2} \frac{d\left(\frac{1}{x}\right)}{\sqrt{1-\frac{1}{x^{2}}}}=-\left.\arcsin \frac{1}{x}\right|_{\sqrt{2}} ^{2}= \\
= & -\arcsin \frac{1}{2}+\arcsin \frac{1}{\sqrt{2}}=-\frac{\pi}{6}+\frac{\pi}{4}=-\frac{2 \pi}{12}+\frac{3 \pi}{12}=\frac{\pi}{12}
\end{aligned}
$$

Source — "http://pluspi.org/wiki/index.php/?%?%?%?%?%?%?%?%?%?%?%?_ %D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD %D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_4-19"

Categories: Kuznetsov's Problem Book Integrals Problem 4 | Integrals

- Last edited on this page: 19:02, 3 March 2009.
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## Problem Kuznetsov Integrals 4-20

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