问题及解答

证明:

\[\int_0^1 \frac{x^4(1-x)^4}{1+x^2}\mathrm{d}x=\frac{22}{7}-\pi\]

参考:

知乎大佬是怎么"注意到"这么恐怖的积分的？_哔哩哔哩_bilibili

利用 Sowya 计算 $\frac{x^4(1-x)^4}{1+x^2}$

>> :mode polyn
Switch into polynomial mode.

>> (x^4(1-x)^4)/(1+x^2)
in> (x^4(1-x)^4)/(1+x^2)

out>
 quotient> q(x) = x^6-4x^5+5x^4-4x^2+4
remainder> r(x) = -4

x^6-4x^5+5x^4-4x^2+4
------------------------

即

\[
\frac{x^4(1-x)^4}{1+x^2}=x^6-4x^5+5x^4-4x^2+4+\frac{-4}{1+x^2}
\]

因此,

\[
\int_0^1\frac{x^4(1-x)^4}{1+x^2}\mathrm{d}x=\int_0^1(x^6-4x^5+5x^4-4x^2+4)\mathrm{d}x+\int_0^1\frac{-4}{1+x^2}\mathrm{d}x
\]