## Problem Statement

Calculate the definite integral:

$$
\int_{1}^{e} \frac{1+\ln x}{x} d x
$$
## Solution

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

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_4-18" Categories: Kuznetsov's Problem Book Integrals Problem 4 | Integrals

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## Problem Kuznetsov Integrals 4-19

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