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Evaluate I = \int_0^1 \frac{x^4(1-x)^4}{1+x^2}\,dx. It can be written…
Question
Evaluate I = \int_0^1 \frac{x^4(1-x)^4}{1+x^2}\,dx. It can be written as I = \frac{a}{b} - \pi, where a/b is a positive rational in lowest terms. Find a+b.
✓ Verified answer: 29checked by our engine — not a guess
Step-by-step solution
Expand the numerator and do polynomial long division by (1+x^2).
One finds x^4(1-x)^4 = (1+x^2)\,q(x) + 4, where q(x)=x^6-4x^5+5x^4-4x^2+4.
Thus the integrand equals q(x) + 4/(1+x^2).
Integrating q(x) on [0,1]: \int_0^1 (x^6-4x^5+5x^4-4x^2+4)dx = 1/7 - 4/6 + 5/5 - 4/3 + 4 = 1/7 - 2/3 + 1 - 4/3 + 4 = 22/7.
And \int_0^1 4/(1+x^2)dx = 4\cdot\arctan 1 = \pi.
Hence I = 22/7 - \pi.
So a=22, b=7 and a+b = 29.
Final answer29
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