## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{1} \frac{x^{3}+x}{x^{4}+1} d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{1} \frac{x^{3}+x}{x^{4}+1} d x=\int_{0}^{1} \frac{x^{3}}{x^{4}+1} d x+\int \frac{x}{x^{4}+1} d x= \\
& =\frac{1}{4} \cdot \int_{0}^{1} \frac{1}{x^{4}+1} d\left(x^{4}+1\right)+\frac{1}{2} \cdot \int_{0}^{1} \frac{1}{x^{4}+1} d\left(x^{2}\right)= \\
& =\frac{1}{4} \cdot \ln \left|x^{4}+1\right|_{0}^{1}+\left.\frac{1}{2} \cdot \arctan x^{2}\right|_{0} ^{1}= \\
& =\frac{1}{4} \cdot \ln \left|1^{4}+1\right|-\frac{1}{4} \cdot \ln \left|0^{4}+1\right|+\frac{1}{2} \cdot \arctan 1^{2}-\frac{1}{2} \cdot \arctan 0^{2}= \\
& =\frac{1}{4} \cdot \ln 2-\frac{1}{4} \cdot \ln 1+\frac{1}{2} \cdot \arctan 1-\frac{1}{2} \cdot \arctan 0= \\
& =\frac{\ln 2}{4}+\frac{1}{2} \cdot \frac{\pi}{4}=\frac{2 \ln 2+\pi}{8}=\frac{\ln 4+\pi}{8}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_4-29$ » Categories: Kuznetsov's Problem Book Integrals Problem 4 | Integrals

- Last edited: 22:22, March 3, 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 4-30

## Material from PlusPi