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$\displaystyle\int_0^1 \dfrac{x^4(1-x)^4}{1 + x^2}\, dx = ?$ (Putnam — pinpoints $22/7 - \pi$)
A$\pi/4$
B$22/7 - \pi$
C$\ln 2$
D$1/7$
Answer & Solution
Correct answer: B. $22/7 - \pi$
Polynomial-divide $x^4(1-x)^4 = x^8 - 4x^7 + 6x^6 - 4x^5 + x^4$. Divide by $1 + x^2$: quotient $x^6 - 4x^5 + 5x^4 - 4x^2 + 4$ with remainder $-4$. So integrand = quotient $- 4/(1+x^2)$. Integrate $0$ to $1$: $1/7 - 4/6 + 1 - 4/3 + 4 - 4\cdot\pi/4 = 1/7 - 2/3 + 1 - 4/3 + 4 - \pi = 22/7 - \pi$.
Related questions
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