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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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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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  • $\begingroup$ reduces to $\frac{22}{7}-\pi.$ $\endgroup$ – Maverick Jan 5 '17 at 13:26
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Hint:

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

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    $\begingroup$ Sir it is $(1-x)^4$. $\endgroup$ – Rohan Jan 5 '17 at 10:37
  • $\begingroup$ Oops! I misread the formula. Fixed. Thanks for pointing it. $\endgroup$ – Bernard Jan 5 '17 at 10:41

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