Integrate: ∫x^4[1 - x]^4 / [1 + x^2]dx,from x=0 to 1
\(\displaystyle\int_0^1\;\frac{x^4(1-x)^4}{1-x^2}\;dx\\ =\displaystyle\int_0^1\;\frac{x^4(1-x)^4}{(1-x)(1+x)}\;dx\\ =-\displaystyle\int_0^1\;\frac{x^4(x-1)^3}{x+1}\;dx\\ =-\displaystyle\int_0^1\;\frac{x^4(x^3-3x^2+3x-1)}{x+1}\;dx\\ =-\displaystyle\int_0^1\;\frac{x^7-3x^6+3x^5-x^4}{x+1}\;dx\\ \text{Do the algebraic division}\\ =-\displaystyle\int_0^1\;x^6-4x^5+7x^4-6x^3+6x^2-6x+6-\frac{6}{x+1}\;dx\\ \)
\( =-\displaystyle\int_0^1\;x^6-4x^5+7x^4-6x^3+6x^2-6x+6-\frac{6}{x+1}\;dx\\ =-\left[\; \frac{x^7}{7}-\frac{4x^6}{6}+\frac{7x^5}{5}-\frac{6x^4}{4}+\frac{6x^3}{3}-\frac{6x^2}{2}+6x-6ln(x+1)\;\right]_0^1\\ =-\left[\; \frac{x^7}{7}-\frac{2x^6}{3}+\frac{7x^5}{5}-\frac{3x^4}{2}+2x^3-3x^2+6x-6ln(x+1)\;\right]_0^1\\ =-\left[\; \frac{1}{7}-\frac{2}{3}+\frac{7}{5}-\frac{3}{2}+2-3+6-6ln(2)\;\right]\\ =-\left[\; \frac{1}{7}-\frac{2}{3}+\frac{7}{5}-\frac{3}{2}+5-6ln(2)\;\right]\)
1/7-2/3+7/5-3/2+5 = 4.3761904761904762 = 919/210
\(=6ln2-\frac{919}{210}\)
-4.3761904761904762+6*log(2) = -2.570010502206589
This is not correct, I have made some silly mistake somewhere, but the method is correct :)
Wolfram Alpha says the answer is approx 0.0023
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