Mathematics - Definite Integration Question with Solution | TestHub

Let $I={\int }_0^1\frac{x^4(1-x{)}^4}{1+x^2}dx$ $$\begin{array}{rl}& ={\int }_0^1\frac{\left(x^4-1\right)(1-x{)}^4+(1-x{)}^4}{\left(1+x^2\right)}dx\\ & ={\int }_0^1\left(x^2-1\right)(1-x{)}^{4}dx\\ & +{\int }_0^1\frac{{\left(1+x^2-2x\right)}^2}{\left(1+x^2\right)}dx\\ & ={\int }_0^1\{\left(x^2-1\right)(1-x{)}^4+\left(1+x^2\right)-4x\\ & +\frac{4x^2}{\left(1+x^2\right)}\}dx\\ & ={\int }_0^1(\left(x^2-1\right)(1-x{)}^4+\left(1+x^2\right)-4x\\ & +4-\frac{4}{1-x^2})dx\\ & ={\int }_0^1(x^6-4x^5+5x^4-4x^2\\ & +4-\frac{4}{1+x^2})dx\\ & =[\frac{x^7}{7}-\frac{4x^6}{6}+\frac{5x^5}{5}\\ & \end{array}$$ $$\begin{array}{rl}& {-\frac{4x^3}{3}+4x-4{\tan }^{-1}x]}_0^1\\ =& \frac{1}{7}-\frac{4}{6}+\frac{5}{5}-\frac{4}{3}+4-4\left(\frac{\pi }{4}-0\right)\\ =& \frac{22}{7}-\pi \end{array}$$