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															Solution
							\[\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=?\]
						
				
							The integral is well defined. Next, let’s use the binomial theorem						
				
							\[(1-x)^{4}=x^{4}-4 x^{3}+6 x^{2}-4 x+1\]						
				
							\[x^{4}(1-x)^{4}=x^{8}-4 x^{7}+6 x^{6}-4 x^{5}+x^{4}\]						
				
							Now, let’s use polynomial division, we get						
				
							\[\frac{x^{4}(1-x)^{4}}{1+x^{2}}=x^{6}-4 x^{5}+5 x^{4}-4 x^{2}+4-\frac{4}{x^{2}+1}\]						
				
							\[\Rightarrow \int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=\int_{0}^{1}\left(x^{6}-4 x^{5}+5 x^{4}-4 x^{2}+4-\frac{4}{x^{2}+1}\right) d x\]						
				
							\[=\frac{x^{7}}{7}-\frac{4 x^{6}}{6}+x^{5}-\frac{4 x^{3}}{3}+4 x-\left.4 \arctan (x)\right|_{0} ^{1}\]						
				
							\[=\frac{1}{7}-\frac{4}{6}+1-\frac{4}{3}+4-4\left(\frac{\pi}{4}\right)-(0-0+0-0+0-4(0))\]						
				
							\[=\frac{22}{7}-\pi\]						
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