## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x-(\operatorname{arctg} x)^{4}}{1+x^{2}} d x
$$

Note: In the context of this problem, $\operatorname{arctg} x$ is the same as $\arctan x$.
## Solution

$$
\begin{aligned}
& \int \frac{x-(\operatorname{arctg} x)^{4}}{1+x^{2}} d x=\int \frac{x d x}{1+x^{2}}-\int \frac{(\operatorname{arctg} x)^{4}}{1+x^{2}}= \\
& =\frac{1}{2} \int \frac{d\left(1+x^{2}\right)}{1+x^{2}}-\int(\operatorname{arctg} x)^{4} d(\operatorname{arctg} x)=\frac{1}{2} \ln \left(1+x^{2}\right)-\frac{(\operatorname{arctg} x)^{5}}{5}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+3-28$ "

Categories: Kuznetsov's Problem Book Integrals Problem 3 | Integrals

Ukrainian Banner Network

- Last edited on this page: 13:41, 3 March 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 3-29

## Material from Plusi