## Problem Statement

Calculate the definite integral:

$$
\int_{1}^{2} x \cdot \ln ^{2} x d x
$$
## Solution

$$
\int_{1}^{2} x \cdot \ln ^{2} x d x=
$$

Let:

$$
\begin{aligned}
& u=\ln ^{2} x ; d u=2 \ln x \cdot \frac{1}{x} \cdot d x \\
& d v=x \cdot d x ; v=\frac{x^{2}}{2}
\end{aligned}
$$

Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:

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

Let:

$$
\begin{aligned}
& u=\ln x ; d u=\frac{1}{x} d x \\
& d v=x \cdot d x ; v=\frac{x^{2}}{2}
\end{aligned}
$$

Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:

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

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5

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Categories: Kuznetsov's Problem Book Integrals Problem 2 | Integrals

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## Problem Kuznetsov Integrals 2-22

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