## Solution

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

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 \%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB\%D1\%8B_4-20»

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

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Created by GeeTeatoo

## Problem Kuznetsov Integrals 4-21

## Material from PlusPi
## Problem Statement

Calculate the definite integral:

$$
\int_{1}^{e} \frac{x^{2}+\ln x^{2}}{x} d x
$$