Question

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

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

Answer 1

Answer: (A)

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

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

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

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

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

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

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

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

$=\frac{1}{7}-\frac{4}{6}+\frac{5}{5}-\frac{4}{3}+4-4(\frac{\pi }{4}-0)$

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