∫_( 0) ^( 1) log(((1 + x + x^2 + x^3 + x^4 )/(√(1 + (1/x) + (1/x^2 ) + (1/x^3 ) + (1/x^4 ))))) dx


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Question Number 77570 by TawaTawa last updated on 08/Jan/20

$\int_{0}^{1} \log (\frac{1 + x + x^{2} + x^{3} + x^{4}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^{2}} + \frac{1}{x^{3}} + \frac{1}{x^{4}}}}) dx$
	Answered by MJS last updated on 08/Jan/20

	$= \underset{0}{\int^{1}} \ln (x^{2} \sqrt{x^{4} + x^{3} + x^{2} + x + 1}) d x =$ $= 2 \underset{0}{\int^{1}} \ln x d x + \frac{1}{2} \underset{0}{\int^{1}} \ln (x^{4} + x^{3} + x^{2} + x + 1) d x$ $2 \underset{0}{\int^{1}} \ln x d x = 2 {[x (\ln x - 1)]}_{0}^{1} =$ $[\lim_{x \to 0} x (\ln x - 1) = 0]$ $= - 2$ $\frac{1}{2} \underset{0}{\int^{1}} \ln (x^{4} + x^{3} + x^{2} + x + 1) d x =$ $[α_{j} = e^{i \frac{2 π j}{5}}; j = 1, 2, 3, 4]$ $= \frac{1}{2} \underset{0}{\int^{1}} \ln (\underset{j = 1}{\prod^{4}} (x - α_{j})) d x =$ $= \frac{1}{2} \underset{j = 1}{\sum^{4}} (\underset{0}{\int^{1}} \ln (x - α_{j}) d x) =$ $= \frac{1}{2} {[\underset{j = 1}{\sum^{4}} ((x - α_{j}) \ln (x - α_{j})) - 2 x]}_{0}^{1} = . . .$ $. . . = \frac{\sqrt{5}}{4} \ln \frac{1 + \sqrt{5}}{2} + \frac{5}{8} \ln 5 + \frac{\sqrt{25 + 10 \sqrt{5}}}{20} π - 2$ $\Rightarrow$ $answer is \frac{\sqrt{5}}{4} \ln \frac{1 + \sqrt{5}}{2} + \frac{5}{8} \ln 5 + \frac{\sqrt{25 + 10 \sqrt{5}}}{20} π - 4$
	Commented by TawaTawa last updated on 08/Jan/20

	$God bless you sir .$
	Commented by TawaTawa last updated on 12/Jan/20

	$Please sir . How did you get : e^{i \frac{2 π j}{5}}; j = 1, 2, 3, 4$
	Commented by MJS last updated on 12/Jan/20

	$(x^{4} + x^{3} + x^{2} + x + 1) (x - 1) = x^{5} - 1$ $\Rightarrow the roots are the 5^{th} roots of unity$ $= e^{i \frac{2 π j}{5}} with j = 0, 1, 2, 3, 4 but j = 0 \Leftrightarrow x - 1 = 0$
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