Solution: Evaluate the integral ∫₀¹ (x⁴(1−x)⁴)/(1+x²) dx

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Solution

In this exercise, we aim to evaluate the integral:

$$
I = \int_0^1 \frac{x^4 (1 – x)^4}{1 + x^2} \, dx
$$

This integral beautifully combines algebraic expansion, polynomial division, and calculus techniques.
By applying the binomial theorem to expand $(1 – x)^4$ and simplifying the resulting rational expression,
we can integrate each term step by step to reach an elegant result involving $ \pi $.

Such problems are not only common in calculus but also serve as an excellent example of how symbolic manipulation
and trigonometric functions can reveal deep mathematical relationships.

\[\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=?\]

The integral is well defined. Next, let’s use the binomial theorem

\[(1-x)^{4}=x^{4}-4 x^{3}+6 x^{2}-4 x+1\]

\[x^{4}(1-x)^{4}=x^{8}-4 x^{7}+6 x^{6}-4 x^{5}+x^{4}\]

Now, let’s use polynomial division, we get

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

\[\Rightarrow \int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=\int_{0}^{1}\left(x^{6}-4 x^{5}+5 x^{4}-4 x^{2}+4-\frac{4}{x^{2}+1}\right) d x\]

\[=\frac{x^{7}}{7}-\frac{4 x^{6}}{6}+x^{5}-\frac{4 x^{3}}{3}+4 x-\left.4 \arctan (x)\right|_{0} ^{1}\]

\[=\frac{1}{7}-\frac{4}{6}+1-\frac{4}{3}+4-4\left(\frac{\pi}{4}\right)-(0-0+0-0+0-4(0))\]

\[=\frac{22}{7}-\pi\]

This result is fascinating because $ \frac{22}{7} $ is a famous historical approximation of $ \pi $,
yet here it appears naturally as part of the exact result of a definite integral.

This demonstrates the deep interconnection between polynomial algebra and transcendental constants,
and highlights the beauty of exact calculus in uncovering mathematical elegance hidden behind seemingly complex expressions.

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