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. $\frac{8}{105}$
• D. $\frac{71}{15}-\frac{3\pi }{2}$

Answer 1

Answer: (A)

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

Expanding the numerator: $x^4(1-x{)}^4=x^8-4x^7+6x^6-4x^5+x^4$

Dividing this by $1+x^2$ by factorizing the terms and getting the reminder:

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

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

$I=1/7-4/6+5/5-4/3+4/1-{\int }_0^1\frac{4}{1+x^2}$

$I=1/7+3-4{\tan }^{-1}1+4{\tan }^{-1}0$

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