Question

The value(s ) of $\underset{0}{\overset{1}{\int }}\frac{x^4(1-x{)}^4}{(1+x^2)}dx$ 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 $I=\underset{0}{\overset{1}{\int }}\frac{x^4(1-x{)}^4}{1+x^2}dx$
$=\underset{0}{\overset{1}{\int }}\frac{\left(x^4-1\right)(1-x{)}^4+(1-x{)}^4}{\left(1+x^2\right)}dx$
$=\underset{0}{\overset{1}{\int }}\left(x^2-1\right)(1-x{)}^{4}dx+\underset{0}{\overset{1}{\int }}\frac{{\left(1+x^2-2x\right)}^2}{\left(1+x^2\right)}dx$
$=\underset{0}{\overset{1}{\int }}\left\{\left(x^2-1\right)(1-x{)}^4+\left(1+x^2\right)-4x+\frac{4x^2}{\left(1+x^2\right)}\right\}dx$
$=\underset{0}{\overset{1}{\int }}\left(\left(x^2-1\right)(1-x{)}^4+\left(1+x^2\right)-4x+4-\frac{4}{1+x^2}\right)dx$
$=\underset{0}{\overset{1}{\int }}\left(x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}\right)dx$
$={\left[\frac{x^7}{7}-\frac{4x^6}{6}+\frac{5x^5}{5}-\frac{4x^3}{3}+4x-4{\tan }^{-1}x\right]}_0^1$
$=\frac{1}{7}-\frac{4}{6}+\frac{5}{5}-\frac{4}{3}+4-4\left(\frac{\pi }{4}-0\right)=\frac{22}{7}-\pi$