I have no idea how to find this integral:
$$\int_0^1\frac{(x-x^2)^4}{1+x^2}\ dx\ ?$$
Any help would be appreciated. Thanks in advance.
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This is a well known integral because of the startling punch line that you see only when you've finished evaluating it: http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80 – Michael Hardy May 06 '14 at 16:48
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@MichaelHardy Thank you very much. I didn't know that. – Venus May 06 '14 at 16:51
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Here's a related link – gar May 06 '14 at 16:52
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1A post about this integral on MO: Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$?. – Martin Sleziak May 13 '18 at 11:13
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See also: Evaluating $\int_0^1 \frac{x^4 (1-x)^4}{1+x^2} dx$ – Martin Sleziak May 13 '18 at 11:32
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HINT : $$ \begin{align} \int_0^1\frac{(x-x^2)^4}{1+x^2}\ dx&=\int_0^1\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}\ dx\tag1\\ &=\int_0^1\left(x^6-4x^5+5x^4-4x^2+4-\frac4{1+x^2}\right)\ dx.\tag2\\ \end{align} $$ See binomial theorem for $(1)$ and polynomial long division for $(2)$. Also $$ \begin{align} \int_0^1\frac1{1+x^2}\ dx=\frac\pi4 \end{align} $$ by letting $x=\tan\theta$. The answer is $\ \color{blue}{\dfrac{22}7-\pi}$.
Tunk-Fey
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it reads as if the answer is $\pi/4$. Perhaps you should clarify it for the OP (getting rid of the equal sign in front of the integral on the last line. – Christopher K May 06 '14 at 16:51
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