How do you integrate this? $\int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}} d x$ $int_0^1(x^4(1-x)^4)/(1+x^2)dx$

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How do you integrate this? $\int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}} d x$ $int_0^1(x^4(1-x)^4)/(1+x^2)dx$

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Answer 1 · 2018-03-17T15:51:06

$\int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}} d x = \frac{22}{7} - π$ $int_0^1(x^4(1-x)^4)/(1+x^2)dx=22/7-pi$

Explanation:

Let

$I = \int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}} d x$ $I=int_0^1(x^4(1-x)^4)/(1+x^2)dx$

Expand the numerator:

$I = \int_{0}^{1} \frac{x^{8} - 4 x^{7} + 6 x^{6} - 4 x^{5} + x^{4}}{x^{2} + 1} d x$ $I=int_0^1(x^8-4x^7+6x^6-4x^5+x^4)/(x^2+1)dx$

Apply long division:

$I = \int_{0}^{1} (x^{6} - 4 x^{5} + 5 x^{4} - 4 x^{2} - \frac{4}{x^{2} + 1} + 4) d x$ $I=int_0^1(x^6-4x^5+5x^4-4x^2-4/(x^2+1)+4)dx$

Integrate directly:

$I = {[\frac{1}{7} x^{7} - \frac{2}{3} x^{6} + x^{5} - \frac{4}{3} x^{3} - 4 {tan}^{- 1} x + 4 x]}_{0}^{7}$ $I=[1/7x^7-2/3x^6+x^5-4/3x^3-4tan^(-1)x+4x]_0^1$

Hence

$I = \frac{22}{7} - π$ $I=22/7-pi$

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How do you integrate this? $\int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}} d x$ $int_0^1(x^4(1-x)^4)/(1+x^2)dx$

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Explanation:

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Explanation:

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How do you integrate this? $\int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}} d x$ $int_0^1(x^4(1-x)^4)/(1+x^2)dx$