## Problem Statement

Calculate the indefinite integral:

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

$$
\int \frac{d x}{x \sqrt{x^{2}-1}}=\int \frac{d x}{x \cdot|x| \cdot \sqrt{1-\frac{1}{x^{2}}}}=
$$

First, consider the case $x>0$:

$$
=\int \frac{d x}{x^{2} \sqrt{1-\frac{1}{x^{2}}}}=
$$

Substitution:

$$
\begin{aligned}
& y=\frac{1}{x} \Rightarrow y^{2}=\frac{1}{x^{2}} \\
& d y=-\frac{1}{x^{2}} d x \Rightarrow \frac{1}{x^{2}} d x=-d y
\end{aligned}
$$

We get:

$$
=\int \frac{-d y}{\sqrt{1-y^{2}}}=-\int \frac{d y}{\sqrt{1-y^{2}}}=-\arcsin y+C=
$$

Reverse substitution:

$$
y=\frac{1}{x}
$$

We get:

$$
=-\arcsin \frac{1}{x}+C
$$

In the case $x<0$, we get:

$$
=\arcsin \frac{1}{x}-C
$$

Since the function $\arcsin x$ is odd, the general solution can be written as:

$$
=-\arcsin \frac{1}{|x|}+C \text { (the sign of the integration constant does not affect the generality) }
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+3-3$ »

Categories: Kuznetsov's Problem Book Integrals Problem $3 \mid$ Integrals

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## Problem Kuznetsov Integrals 3-4

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