## Task Condition

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

$$
y=\frac{x}{\left(x^{2}+1\right)^{2}}, y=0, x=1
$$
## Solution

Find the abscissas of the points of intersection of the graphs of the functions

$$
\begin{gathered}
y=\frac{x}{\left(x^{2}+1\right)^{2}}, y=0 \\
\frac{x}{\left(x^{2}+1\right)^{2}}=0 \\
x=0
\end{gathered}
$$

In the case of $y=\frac{x}{\left(x^{2}+1\right)^{2}}, x=1$ it is obvious:

$$
x=1
$$

Calculate the area:

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-43.jpg?height=1019&width=1037&top_left_y=987&top_left_x=909)

$$
\begin{aligned}
& S=\int_{0}^{1}\left(\frac{x}{\left(x^{2}+1\right)^{2}}-0\right) d x=\int_{0}^{1} \frac{x d x}{\left(x^{2}+1\right)^{2}}= \\
& =\frac{1}{2} \cdot \int_{0}^{1} \frac{d\left(x^{2}+1\right)}{\left(x^{2}+1\right)^{2}}=\left.\frac{1}{2} \cdot \frac{1}{-1} \cdot\left(x^{2}+1\right)^{-1}\right|_{0} ^{1}= \\
& =-\frac{1}{2} \cdot\left(\left(1^{2}+1\right)^{-1}-\left(0^{2}+1\right)^{-1}\right)=-\frac{1}{2} \cdot\left(\frac{1}{2}-1\right)= \\
& =-\frac{1}{2} \cdot\left(-\frac{1}{2}\right)=\frac{1}{4}
\end{aligned}
$$

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## Problem Kuznetsov Integrals 14-24

## Material from PlusPi