## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{1+\ln (x-1)}{x-1} d x
$$
## Solution

$$
\begin{aligned}
& \int \frac{1+\ln (x-1)}{x-1} d x=\int \frac{d x}{x-1}+\int \frac{\ln (x-1)}{x-1} d x= \\
& =\int \frac{d(x-1)}{x-1}+\int \ln (x-1) d(\ln (x-1))= \\
& =\ln (x-1)+\frac{1}{2} \cdot \ln ^{2}(x-1)+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_3-16$ "

Categories: Kuznetsov's Problem Book Integrals Problem 3 | Integrals

- Last edited: 18:13, 27 February 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 3-17

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