## Problem Statement

Calculate the definite integral:

$$
\int_{e+1}^{e^{2}+1} \frac{1+\ln (x-1)}{x-1} d x
$$
## Solution

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+4-1$ » Categories: Kuznetsov's Problem Book Integrals Problem 4 | Integrals

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## Problem Kuznetsov Integrals 4-2

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