## Problem Statement

Calculate the area of the figure bounded by the graphs of the functions:

$$
y=\frac{1}{x \sqrt{1+\ln x}}, y=0, x=1, x=e^{3}
$$
## Solution

Answer: 2

We construct the graphs:

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-17.jpg?height=1059&width=1585&top_left_y=1047&top_left_x=218)

We obtain a figure bounded above by the curve $\frac{1}{x \cdot \sqrt{1+\ln x}}$, on the left by the line $x=1$, on the right by the line $x=e^{3}$, and below by $y=0$.

Thus, the area of the desired figure will be equal to the integral from 1 to $e^{3}$ of the function $\frac{1}{x \cdot \sqrt{1+\ln x}}$. We take the integral:

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

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 $\% \mathrm{D} 1 \% 84 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-9$ »

Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

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## Problem Kuznetsov Integrals 14-10

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