## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{1} \frac{x^{3} d x}{\left(x^{2}+1\right)^{2}}
$$
## Solution

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

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_4-22" Categories: Kuznetsov's Problem Book Integrals Problem 4 | Integrals

- Last edited: 19:46, March 3, 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 4-23

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