## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{2}+\ln x^{2}}{x} d x
$$
## Solution

$$
\int \frac{x^{2}+\ln x^{2}}{x} d x=\int x \cdot d x+\int \frac{\ln x^{2}}{x} d x=\frac{x^{2}}{2}+\int \frac{\ln x^{2}}{x} d x=
$$

Substitution:

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

We get:

$$
=\frac{x^{2}}{2}+\int \frac{y \cdot d y}{2}=\frac{x^{2}}{2}+\frac{y^{2}}{4}+C=
$$

After reverse substitution:

$$
\begin{aligned}
& =\frac{x^{2}}{2}+\frac{\left(\ln x^{2}\right)^{2}}{4}+C=\frac{x^{2}}{2}+\frac{(2 \ln x)^{2}}{4}+C= \\
& =\frac{x^{2}}{2}+\frac{4 \ln ^{2} x}{4}+C=\frac{x^{2}}{2}+\ln ^{2} x+C
\end{aligned}
$$

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

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

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## Problem Kuznetsov Integrals $3-5$

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