## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{1} \frac{x}{\sqrt{x^{4}+x^{2}+1}} d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{1} \frac{x}{\sqrt{x^{4}+x^{2}+1}} d x=\int_{0}^{1} \frac{\frac{1}{2} \cdot d\left(x^{2}+\frac{1}{2}\right)}{\sqrt{\left(x^{2}+\frac{1}{2}\right)^{2}+\frac{3}{4}}}=\frac{1}{2} \cdot \ln \left|x^{2}+\frac{1}{2}+\sqrt{\left(x^{2}+\frac{1}{2}\right)^{2}+\frac{3}{4}}\right|_{0}^{1}= \\
& =\frac{1}{2} \cdot \ln \left|1^{2}+\frac{1}{2}+\sqrt{\left(1^{2}+\frac{1}{2}\right)^{2}+\frac{3}{4}}\right|-\frac{1}{2} \cdot \ln \left|0^{2}+\frac{1}{2}+\sqrt{\left(0^{2}+\frac{1}{2}\right)^{2}+\frac{3}{4}}\right|= \\
& =\frac{1}{2} \cdot \ln \left|\frac{3}{2}+\sqrt{\frac{9}{4}}+\frac{3}{4}\right|-\frac{1}{2} \cdot \ln \left|\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{3}{4}}\right|=\frac{1}{2} \cdot \ln \left|\frac{3}{2}+\sqrt{3}\right|-\frac{1}{2} \cdot \ln \left|\frac{1}{2}+1\right|= \\
& =\ln \sqrt{\frac{3+2 \sqrt{3}}{2}}-\ln \sqrt{\frac{3}{2}}=\ln \frac{\sqrt{\frac{3+2 \sqrt{3}}{2}}}{\sqrt{\frac{3}{2}}}=\ln \sqrt{\frac{3+2 \sqrt{3}}{3}}
\end{aligned}
$$

Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�

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$\% \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} \_4-21 »"

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

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## Problem Kuznetsov Integrals 4-22

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