Question

integration of {x^4(1-x)^4}/(1+x^2) where the limit is from 0 to 1​

Answer 1

refer to the attachment please

the integration sign is not available with me

so I did it in written..

HOPE IT HELPS

Answer 2

Given Question : -

$\bf \:Evaluate \: \sf\int\limits_{0}^{1} \dfrac{ {x}^{4} {(1 - x)}^{4}dx }{1+x^2}$

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$\bf \:\large \red{Identity \: used:- } ✍$

$\bf \: {(x - y)}^{4} = {x}^{4} - {4x}^{3} y + 4 {x}^{2} {y}^{2} - 4x {y}^{3} + {y}^{4}$

$\huge \orange{AηsωeR}✍$

$\bf \:\large \red{Consider} ✍$

$\bf\int\limits_{0}^{1} \dfrac{ {x}^{4} {(1 - x)}^{4}dx }{1+x^2}$

$\sf \: ⟼\bf\int\limits_{0}^{1} \dfrac{ {x}^{4} {(1 - 4x + 6 {x}^{2} - 4 {x}^{3} + x}^{4})dx }{1+x^2}$

$\sf \: ⟼\bf\int\limits_{0}^{1} \dfrac{ {( {x}^{4} - 4 {x}^{5} + 6 {x}^{6} - 4 {x}^{7} + x}^{8})dx }{1+x^2}$

☆ As, degree of numerator > degree of denominator,

☆ So, after applying long division, we get

$\sf \: ⟼\bf\int\limits_{0}^{1}( {x}^{6} - 4 {x}^{5} + {5x}^{4} - {4x}^{2} + 4 - \dfrac{4}{ {x}^{2} + 1 } )dx$

$\bf \:  ⟼ [\dfrac{1}{7} x^7-\dfrac{2}{3} x^6+x^5-\dfrac{4}{3} x^3-4 {tan}^{ - 1}x +4x]_0^1$

$\bf \:  ⟼ \dfrac{1}{7} - \dfrac{2}{3} + 1 - \dfrac{4}{3} + 4 - 4( {tan}^{ - 1} 1)$

$\bf \:  ⟼ \dfrac{1}{7} - (\dfrac{4}{3} + \dfrac{2}{3} ) + 5 - 4 \times \dfrac{\pi}{4}$

$\bf \:  ⟼ \dfrac{1}{7} - 2 + 5 - \pi$

$\bf \:  ⟼ \dfrac{1}{7} + 3 - \pi$

$\bf \:  ⟼ \dfrac{22}{7} - \pi$

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