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We have to solve the following $$\int_0^1 \frac{x^4 (1-x)^4}{1+x^2} dx$$

I tried to substitute $x =\tan m$, but in that I got stuck.

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    https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80#Details_of_evaluation_of_the_integral – 2017-01-07
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    This seems to be rather popular integral: [Finding $\int_0^1{\frac{x^4(1-x)^4}{1+x^2}}dx$](http://math.stackexchange.com/q/129625), [Shortest method for $\int^{1}_{0}\frac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}$](http://math.stackexchange.com/q/1450910), [Creative way for $\int _0 ^1 \frac{x^4 (1-x)^4} {x^2 +1} {\rm d}x$](http://math.stackexchange.com/q/1614676). Found [using Approach0](https://approach0.xyz/search/?q=%24%5Cint_0%5E1%20%5Cfrac%7Bx%5E4%20(1-x)%5E4%7D%7B1%2Bx%5E2%7D%20dx%24&p=1). – 2017-01-07

3 Answers 3

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Hint -

Simplify term by multiplying $x^4(1-x)^2(1-x)^2$ then divide by $1+x^2$. Then easily integrate it.

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Apply long division on the integrand to obtain $$\int_0^1 \left(x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{1+x^2}\right)dx,$$ which is easy to solve.

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    reduces to $\frac{22}{7}-\pi.$ – 2017-01-05
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Hint:

No substitution required here: just divide $x^4(1-x)^4$ by $x^2+1$.

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    Sir it is $(1-x)^4$. – 2017-01-05
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    Oops! I misread the formula. Fixed. Thanks for pointing it. – 2017-01-05