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)

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