∫(dx/(x+(√x))) (2,3)=?


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Question Number 94811 by Abdulrahman last updated on 21/May/20

$\int \frac{dx}{x + \sqrt{x}} (2, 3) = ?$
	Answered by ElOuafi last updated on 21/May/20

	$\int \frac{d x}{x + \sqrt{x}} = \int \frac{d x}{\sqrt{x} (\sqrt{x} + 1)} = \int \frac{\frac{2}{2 \sqrt{x}}}{\sqrt{x} + 1} d x = 2 \int \frac{d (\sqrt{x} + 1)}{\sqrt{x} + 1}$ $p u t u = \sqrt{x} + 1 \Rightarrow= 2 \int \frac{d u}{u}$ $= 2 l n (\sqrt{x} + 1) + c$ $s o \int_{2}^{3} \frac{d x}{x + \sqrt{x}} = 2 l n (\frac{\sqrt{3} + 1}{\sqrt{2} + 1})$
	Commented by Abdulrahman last updated on 21/May/20

	$thanks alot$
	Answered by i jagooll last updated on 21/May/20
	$let (√(x ))=t ⇒x=t^2 ; dx = 2t dt ∫ ((2t dt)/(t^2 +t)) = ∫ ((2 dt)/(t+1)) = 2 ln ∣t+1∣ = 2[ ln ∣1+(√x) ∣ ]_2 ^3 = 2 {ln (1+(√3))−ln(1+(√2)) } = 2 ln (((1+(√3))/(1+(√2))))$
	$let \sqrt{x} = t \Rightarrow x = t^{2}; dx = 2 t dt$ $\int \frac{2 t dt}{t^{2} + t} = \int \frac{2 dt}{t + 1} = 2 \ln ∣ t + 1 ∣$ $= 2 {[\ln ∣ 1 + \sqrt{x} ∣]}_{2}^{3}$ $= 2 {\ln (1 + \sqrt{3}) - \ln (1 + \sqrt{2})}$ $= 2 \ln (\frac{1 + \sqrt{3}}{1 + \sqrt{2}})$
	Commented by Abdulrahman last updated on 21/May/20

	$thanks alot$
	Answered by mathmax by abdo last updated on 21/May/20

	$I = \int_{2}^{3} \frac{dx}{x + \sqrt{x}} we do the changement \sqrt{x} = t \Rightarrow$ $I = \int_{\sqrt{2}}^{\sqrt{3}} \frac{2 tdt}{t^{2} + t} = 2 \int_{\sqrt{2}}^{\sqrt{3}} \frac{dt}{t + 1} = 2 {[\ln (t + 1)]}_{\sqrt{2}}^{\sqrt{3}} = 2 (\ln (1 + \sqrt{3}) - \ln (1 + \sqrt{2}))$
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