## Problem Statement

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

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

The desired area $S$ is:

$S=\int_{1}^{e^{3}} \frac{1}{y \sqrt{\ln y+1}} d y$

We make a substitution of variables:

$t=\ln y$, hence

$d t=\frac{d y}{y}$

When $t=\ln 1 \Rightarrow t=0$ and when

$t=\ln e^{3} \Rightarrow t=3$

Then we get

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-47.jpg?height=1357&width=1168&top_left_y=989&top_left_x=752)

$$
\begin{aligned}
& S=\int_{1}^{e^{3}} \frac{1}{y \sqrt{\ln y+1}} d y= \\
& =\int_{0}^{3} \frac{1}{\sqrt{t+1}} d t=\left.2 \sqrt{1+t}\right|_{0} ^{3}=2 \sqrt{4}-2 \sqrt{1}=2
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-25$ »

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

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## Problem Kuznetsov Integrals 14-26

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