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Question Number 118679 by Algoritm last updated on 19/Oct/20

	Answered by 1549442205PVT last updated on 19/Oct/20

	$\int_{0}^{1} \frac{[2 x] . x}{[2 x] + x^{2}} dx = \int_{0}^{\frac{1}{2}} \frac{[2 x] . x}{[2 x] + x^{2}} dx + \int_{\frac{1}{2}}^{1} \frac{[2 x] . x}{[2 x] + x^{2}} dx$ $= \int_{\frac{1}{2}}^{1} \frac{x}{1 + x^{2}} dx = \frac{1}{2} \int \frac{d (x^{2} + 1)}{x^{2} + 1} = \frac{1}{2} \ln (x^{2} + 1) ∣_{\frac{1}{2}}^{1}$ $= \frac{1}{2} \ln 2 - \ln \sqrt{\frac{5}{4}} = \frac{3}{2} \ln 2 - \ln \sqrt{5} \approx 0.2350$ $Since [2 x] = 0 for 0 ⩽ x < 0.5$ $[2 x] = 1 for 0.5 ⩽ x < 1, [2 x] = 2 for x = 1$
	Commented by Dwaipayan Shikari last updated on 19/Oct/20

	$\frac{3}{2} l o g 2 - \frac{1}{2} l o g 5 = \frac{1}{2} l o g 8 - \frac{1}{2} l o g 5 = \frac{1}{2} l o g (\frac{8}{5})$
	Answered by mathmax by abdo last updated on 19/Oct/20
	$I =∫_0 ^1 ((x[2x])/(x^2 +[2x]))dx ⇒I =_(2x=t) ∫_0 ^2 (t/2)×(([t])/((t^2 /4)+[t]))×(dt/2) =∫_0 ^2 ((t[t])/(t^2 +4[t]))dt ={ ∫_0 ^1 o dt +∫_1 ^2 (t/(t^2 +4))dt} =(1/2)∫_1 ^2 ((2t)/(t^2 +4))dt =(1/2)[ln(t^2 +4)]_1 ^2 =(1/2){ln(8)−ln5}=(1/2)ln((8/5)) answer B is correct$
	$I = \int_{0}^{1} \frac{x [2 x]}{x^{2} + [2 x]} dx \Rightarrow I =_{2 x = t} \int_{0}^{2} \frac{t}{2} \times \frac{[t]}{\frac{t^{2}}{4} + [t]} \times \frac{dt}{2}$ $= \int_{0}^{2} \frac{t [t]}{t^{2} + 4 [t]} dt = {\int_{0}^{1} o dt + \int_{1}^{2} \frac{t}{t^{2} + 4} dt}$ $= \frac{1}{2} \int_{1}^{2} \frac{2 t}{t^{2} + 4} dt = \frac{1}{2} {[\ln (t^{2} + 4)]}_{1}^{2} = \frac{1}{2} {\ln (8) - \ln 5} = \frac{1}{2} \ln (\frac{8}{5})$ $answer B is correct$
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