## Problem Statement

Calculate the definite integral:

$$
\int_{\sqrt{3}}^{\sqrt{8}} \frac{d x}{x \sqrt{x^{2}+1}}
$$
## Solution

$$
\begin{aligned}
& \int_{\sqrt{3}}^{\sqrt{8}} \frac{1}{x \sqrt{x^{2}+1}} d x=\int_{\sqrt{3}}^{\sqrt{8}} \frac{1}{x^{2} \cdot \frac{1}{x} \sqrt{x^{2}+1}} d x=\int_{\sqrt{3}}^{\sqrt{8}} \frac{1}{x^{2} \sqrt{1+\frac{1}{x^{2}}}} d x= \\
& =-\int_{\sqrt{3}}^{\sqrt{3}} \frac{1}{\sqrt{1+\frac{1}{x^{2}}}} d\left(\frac{1}{x}\right)=-\left.\ln \left|\frac{1}{x}+\sqrt{\frac{1}{x^{2}}+1}\right|\right|_{\sqrt{3}} ^{\sqrt{8}}= \\
& =-\ln \left|\frac{1}{\sqrt{8}}+\sqrt{\frac{1}{(\sqrt{8})^{2}}+1}\right|+\ln \left|\frac{1}{\sqrt{3}}+\sqrt{\frac{1}{(\sqrt{3})^{2}}+1}\right|= \\
& =-\ln \left|\frac{1+\sqrt{9}}{\sqrt{8}}\right|+\ln \left|\frac{1+\sqrt{4}}{\sqrt{3}}\right|=-\ln \left|\frac{4}{2 \sqrt{2}}\right|+\ln \left|\frac{3}{\sqrt{3}}\right|= \\
& =-\ln \sqrt{2}+\ln \sqrt{3}=\ln \sqrt{\frac{3}{2}}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+4-17$ »

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

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## Problem Kuznetsov Integrals 4-18

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