## Problem Statement

Calculate the definite integral:

$$
\int_{-1}^{0}(x+2)^{3} \cdot \ln ^{2}(x+2) d x
$$
## Solution

$$
\int_{-1}^{0}(x+2)^{3} \cdot \ln ^{2}(x+2) d x=
$$

Let's denote:

$$
\begin{aligned}
& u=\ln ^{2}(x+2) ; d u=2 \ln (x+2) \cdot \frac{1}{x+2} \cdot d x \\
& d v=(x+2)^{3} \cdot d x ; v=\frac{(x+2)^{4}}{4}
\end{aligned}
$$

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

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

Let's denote:

$$
\begin{aligned}
& u=\ln (x+2) ; d u=\frac{1}{x+2} d x \\
& d v=(x+2)^{3} \cdot d x ; v=\frac{(x+2)^{4}}{4}
\end{aligned}
$$

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

$$
\begin{aligned}
& =4 \ln ^{2} 2-\frac{1}{2} \cdot\left(\left.\ln (x+2) \cdot \frac{(x+2)^{4}}{4}\right|_{-1} ^{0}-\int_{-1}^{0} \frac{(x+2)^{4}}{4} \cdot \frac{1}{x+2} d x\right)= \\
& =4 \ln ^{2} 2-\frac{1}{2} \cdot\left(\ln (0+2) \cdot \frac{(0+2)^{4}}{4}-\ln (-1+2) \cdot \frac{(-1+2)^{4}}{4}-\frac{1}{4} \int_{-1}^{0}(x+2)^{3} \cdot d x\right)= \\
& =4 \ln ^{2} 2-\frac{1}{2} \cdot\left(4 \ln 2-0 \cdot \frac{1^{4}}{4}-\left.\frac{1}{4} \cdot \frac{(x+2)^{4}}{4}\right|_{-1} ^{0}\right)=4 \ln ^{2} 2-\frac{1}{2} \cdot\left(4 \ln 2-\frac{(0+2)^{4}}{16}+\frac{(-1+2)^{4}}{16}\right)=
\end{aligned}
$$

$$
\begin{aligned}
& =4 \ln ^{2} 2-\frac{1}{2} \cdot\left(4 \ln 2-1+\frac{1}{16}\right)=4 \ln ^{2} 2-2 \ln 2+\frac{1}{2}-\frac{1}{32}= \\
& =4 \ln ^{2} 2-2 \ln 2+\frac{15}{32}
\end{aligned}
$$

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

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

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## Problem Kuznetsov Integrals 2-27

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