definite integral.....

The answer should be A.

This is a definite integral question but I could not make it out with any property. Just expand the term (1-x)⁴ and make factors of (1+x²).

yes vivek you are right

\hspace{-16}\int_{0}^{1}\frac{x^4.(1-x^4)}{1+x^2}dx$\\\\\\ Put $x=\tan\theta\Leftrightarrow dx=\sec^2\thetad\theta$ and Changing Limit\\\\ $\int_{0}^{\frac{\pi}{4}}\frac{\tan^4\theta.(1-\tan\theta)^4}{\sec^2\theta}.\sec^2\theta d\theta$\\\\\\ $\int_{0}^{\frac{\pi}{4}}\tan^4\theta.(\tan^4\theta-4\tan^3\theta+6\tan^2\theta-4\tan\theta+1)d\theta$\\\\ $\int_{0}^{\frac{\pi}{4}}\left(\tan^8\theta-4\tan^7\theta+6\tan^6\theta-4\tan^5\theta+\tan^4\theta\right)d\theta$\\\\ Now Using The formula $\boxed{I_{n}+I_{n-2}=\frac{1}{n-1}}$\\\\ Where $\boxed{I_{n}=\int_{0}^{\frac{\pi}{4}}\tan^n xdx\;, n\in\mathbb{N}}$\\\\ So Given Integral can be Written as \\\\ $\int_{0}^{\frac{\pi}{4}}\left\{\left(\tan^8 \theta+\tan^6 \theta\right)+\left(\tan^6 \theta+\tan^4 \theta\right)-4.\left(\tan^7 \theta+\tan^5 \theta\right)+4\tan^6\theta\right\} d\theta$\\\\\\

\hspace{-16}=\frac{1}{7}+\frac{1}{5}-4.\frac{1}{6}+4.\int_{0}^{\frac{\pi}{4}}\left(\tan^6 \theta +\tan^4\theta\right)d\theta-4.\int_{0}^{\frac{\pi}{4}}\left(\tan^4\theta+\tan^2\theta\right)d\theta$\\\\\\ \hspace{56} \;\;\; $4.\int_{0}^{\frac{\pi}{4}}\tan^2\theta d\theta$\\\\\\ $=\frac{1}{7}+\frac{1}{5}-\frac{2}{3}+4.\frac{1}{5}-4.\frac{1}{3}+4.\left(1-\frac{\pi}{4}\right)$\\\\\\ $=\frac{22}{7}-\pi$

This was an IIT-JEE 2010 question .I had the solution but thought that it must have a smaller solution.

BTW thanks, Vivek and Man111 for a nice solution.

2010, Maths paper was quite lengthy I presume.