## Problem Statement

Calculate the definite integral:

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

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

Let's denote:

$$
\begin{aligned}
& u=\ln ^{2} x ; d u=2 \ln x \cdot \frac{1}{x} \cdot d x \\
& d v=\sqrt{x} \cdot d x ; v=\frac{\sqrt{x^{3}}}{\left(\frac{3}{2}\right)}=\frac{2 \sqrt{x^{3}}}{3}
\end{aligned}
$$

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

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

Let's denote:

$$
\begin{aligned}
& u=\ln x ; d u=\frac{1}{x} d x \\
& d v=\sqrt{x} \cdot d x ; v=\frac{2 \sqrt{x^{3}}}{3}
\end{aligned}
$$

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

$$
\begin{aligned}
& =\frac{2 e \sqrt{e}}{3}-\frac{4}{3} \cdot\left(\left.\ln x \cdot \frac{2 \sqrt{x^{3}}}{3}\right|_{1} ^{e}-\int_{1}^{e} \frac{2 \sqrt{x^{3}}}{3} \cdot \frac{1}{x} d x\right)=\frac{2 e \sqrt{e}}{3}-\frac{4}{3} \cdot\left(\ln e \cdot \frac{2 \sqrt{e^{3}}}{3}-\ln 1 \cdot \frac{2 \sqrt{1^{3}}}{3}-\frac{2}{3}\right. \\
& =\frac{2 e \sqrt{e}}{3}-\frac{4}{3} \cdot\left(\frac{2 e \sqrt{e}}{3}-0 \cdot \frac{2 \sqrt{1^{3}}}{3}-\left.\frac{2}{3} \cdot \frac{2 \sqrt{x^{3}}}{3}\right|_{1} ^{e}\right)=\frac{2 e \sqrt{e}}{3}-\frac{4}{3} \cdot\left(\frac{2 e \sqrt{e}}{3}-\frac{4 \sqrt{e^{3}}}{9}+\frac{4 \sqrt{1^{3}}}{9}\right)= \\
& =\frac{2 e \sqrt{e}}{3}-\frac{4}{3} \cdot\left(\frac{6 e \sqrt{e}}{9}-\frac{4 e \sqrt{e}}{9}+\frac{4}{9}\right)=\frac{2 e \sqrt{e}}{3}-\frac{4}{3} \cdot\left(\frac{2 e \sqrt{e}}{9}+\frac{4}{9}\right)=
\end{aligned}
$$

$$
=\frac{18 e \sqrt{e}}{27}-\frac{8 e \sqrt{e}}{27}-\frac{16}{27}=\frac{10 e \sqrt{e}-16}{27}
$$

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\%D0\%B3 $\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_2-28$ »

Categories: Kuznetsov's Problem Book Integrals Problem $2 \mid$ Integrals

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## Problem Kuznetsov Integrals 2-29

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