Question

Find the value(s) of

$\int_{0}^{1}\frac{{x}^{4}{(1-x)}^{4}}{(1+{x}^{2})}dx$​

Answer 1

Answer:

Step-by-step explanation:

brainliest please

Answer 2

ANSWER

$- \pi + \dfrac{22}{7}$

Step By Step Explanation

We have >

$\int_{0}^{1}\dfrac{{x}^{4}.{(1-x)}^{4} }{(1+{x}^{2})}dx$

$\implies \int_{0}^{1}\dfrac{(1 + x^{2} - 2x)^{2}.x^{4} }{1+x^{2} }dx$

$\implies \int_{0}^{1}\dfrac{ (1 + x^{2})^{2} + 4x^{2} - 4x(1 + x^{2})x^{4} }{1+ x^{2} }dx$

$\implies \int_{0}^{1}\dfrac{ (1 + x^{4} + 2x^{2} + 4x^{2} - 4x - 4x^{3} )x^{4} }{1+ x^{2} }dx$

$\implies \int_{0}^{1}\dfrac{ (1 + x^{4} - 4x^{3} + 6x^{2} - 4x)x^{4} }{1+ x^{2} }dx$

$\implies \int_{0}^{1}\dfrac{ x^{8} - 4x^{7} + 6x^{6} - 4x^{5} + x^{4} }{1+ x^{2} }dx$

By using long division method we get > >

$\implies \int_{0}^{1}[\dfrac{ -4 } {1 + x^{2} } + x^{6} - 4x^{5} +5x^{4} - 4x^{2} + 4]dx$

$\implies [ -4tan^{-1}x + \dfrac{x^{7} }{7} - \dfrac{4x^{6} }{6} + \dfrac{5x^{5} }{5} - \dfrac{4x^{3} }{3} + 4x]_{0}^{1}$

$\implies -4 \times \dfrac{\pi}{4} + \dfrac{1}{7} - \dfrac{4}{6} + \dfrac{5}{5} - \dfrac{4}{3} + 4$

$-\pi + \dfrac{22}{7}$