Question

The value (s) of${\int }_0^1\frac{x^4{\left(1-x\right)}^4}{1+x^2}dx$ is (are)

• A. $\frac{22}{7}-\mathrm{\pi }$
• B. $\frac{2}{105}$
• C. $0$
• D. $\frac{71}{15}-\frac{3\mathrm{\pi }}{2}$

Answer 1

The correct option is A

$\frac{22}{7}-\mathrm{\pi }$

Explanation for the correct option:

Given, ${\int }_0^1\frac{x^4{\left(1-x\right)}^4}{1+x^2}dx$

Simplifying,

$\begin{array}{rcl}{\int }_0^1\frac{x^4{\left(1-x\right)}^4}{1+x^2}dx& =& {\int }_0^1\frac{\left(x^4-1\right){\left(1-x\right)}^4+{\left(1-x\right)}^4}{1+x^2}dx\\ & =& {\int }_0^1\frac{\left(x^2+1\right)\left(x^2-1\right){\left(1-x\right)}^4}{1+x^2}dx+{\int }_0^1\frac{{\left(1-x\right)}^2{\left(1-x\right)}^2}{1+x^2}dx\\ & =& {\int }_0^1\left(x^2-1\right)\left(1-x^4\right)dx+{\int }_0^1\frac{{\left(1+x^2-2x\right)}^2}{1+x^2}dx\\ & =& {\int }_0^1\left[\left(x^2-1\right){\left(1-x\right)}^4+\left(1+x^2\right)+\frac{4x}{1+x^2}-4x\right]dx\\ & =& {\int }_0^1{x}^6-4x^5+5x^4-4x^3+4-\frac{4}{1+x^2}dx\\ & =& {\left[\frac{x^7}{7}-\frac{4x^6}{6}+\frac{5x^5}{5}-\frac{4x^3}{3}+4x-4{\tan }^{-1}\left(x\right)\right]}_0^1\\ & =& \frac{1}{7}-\frac{4}{6}+1-\frac{4}{3}+4-4\left(\frac{\mathrm{\pi }}{4}\right)\\ & =& \frac{22}{7}-\mathrm{\pi }\end{array}$

Hence, option A is correct.