## Problem Statement

Calculate the indefinite integral:

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

$$
\begin{aligned}
& \int \frac{x-\frac{1}{x}}{\sqrt{x^{2}+1}} d x=\int \frac{x}{\sqrt{x^{2}+1}} d x-\int \frac{1}{x \sqrt{x^{2}+1}} d x= \\
& =\frac{1}{2} \cdot \int \frac{d\left(x^{2}+1\right)}{\sqrt{x^{2}+1}}-\int \frac{1}{x \sqrt{x^{2}+1}} d x=\frac{1}{2} \cdot 2 \cdot \sqrt{x^{2}+1}-\int \frac{1}{x^{2} \cdot \frac{1}{x} \sqrt{x^{2}+1}} d x= \\
& =\sqrt{x^{2}+1}-\int \frac{1}{x^{2} \sqrt{1+\frac{1}{x^{2}}}} d x=\sqrt{x^{2}+1}+\int \frac{1}{\sqrt{1+\frac{1}{x^{2}}}} d\left(\frac{1}{x}\right)= \\
& =\sqrt{x^{2}+1}+\ln \left|\frac{1}{x}+\sqrt{\frac{1}{x^{2}}+1}\right|+C=\sqrt{x^{2}+1}+\ln \left|\frac{1+\sqrt{x^{2}+1}}{x}\right|+C
\end{aligned}
$$

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 $\% \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} \_3-26 »"

Categories: Kuznetsov's Problem Book Integrals Problem 3 | Integrals

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## Problem Kuznetsov Integrals 3-27

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