Question

Evaluate ${\int }_0^1\frac{x^4{(1-x)}^4}{1+x^2}dx$

• A. $(\pi -\frac{22}{7})$
• B. $(\frac{22}{7}-\pi )$
• C. $(\frac{7}{22}-\pi )$
• D. $(\frac{22}{7}+\pi )$

Answer 1

Answer: (B)

$(\frac{22}{7}-\pi )$

${\int }_0^1\frac{x^4{(1-x)}^4}{1+x^2}dx$=${\int }_0^1\frac{x^4{[1+x^2-2x]}^2}{1+x^2}$

$=={\int }_0^1{x}^4(1+x^2)dx-4{\int }_0^1{x}^{5}dx+4{\int }_0^1\frac{x^6}{1+x^2}dx$

$=[\frac{x^2}{5}+\frac{x^7}{7}]-4{\left[\frac{x^6}{6}\right]}_0^1+4{\int }_0^1\frac{-dx}{1+x^2}+4{\int }_1^0\frac{(x^2+1)(x^4+1-x^2)}{1+x^2}dx$

$=(\frac{1}{5}+\frac{1}{7})-4\left(\frac{1}{6}\right)-4{\left[{\tan }^{-1}x\right]}_0^1+(\frac{1}{5}+1-\frac{1}{3})$

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