## Problem Statement

Calculate the definite integral:

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

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

Let's denote:

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

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

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

Let's denote:

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

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

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

$$
\begin{aligned}
& =9 \ln ^{2} 3-\frac{2}{3} \cdot\left(9 \ln 3-3+\frac{1}{9}\right)=9 \ln ^{2} 3-6 \ln 3+2-\frac{2}{27}= \\
& =9 \ln ^{2} 3-6 \ln 3+1 \frac{25}{27}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\� \%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80 $\% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} 2-27$ »

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

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- Last edited: 09:53, March 23, 2009.
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## Problem Kuznetsov Integrals 2-28

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