Question

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

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

Answer 1

The correct option is A $\frac{22}{7}-\pi$

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

Now,$f\left(x\right)=\frac{x^4{\left(1-x\right)}^4}{1+x^2}=\frac{x^4{\left(1-x\right)}^4{\left(1-x\right)}^2}{1+x^2}$

$\begin{array}{c}\Rightarrow \frac{x^4{\left(1+x^2-2x\right)}^2}{1+x^2}\\ \Rightarrow \frac{\left(1+x^4+4x^2-4x^3-4x\right)x^4}{1+x^2}\\ \Rightarrow \frac{2^8-4x^7+6x^6-4x^5+x^4}{1+x^2}\end{array}$

Now, After division.

$\begin{array}{c}{\int }_0^{1}f\left(x\right)={\int }_0^1\left\{\left(x^6-4x^5+5x^4-4x^4+4\right)-\frac{4}{1+x^2}\right\}\\ \Rightarrow {{\left(\frac{x^7}{7}-\frac{4x^6}{6}+\frac{5x^5}{5}-\frac{4x^2}{3}+4x\right)}^1}_0-4{{\left({\tan }^{-1}x\right)}^1}_0\\ \Rightarrow \left(1/7-2/3+1-\frac{4}{3}+4-4\cdot \pi /4\right)=\frac{22}{7}-\pi \end{array}$

Hence,

Option $A$ is correct answer