prove the convergence of ∫_0 ^1 ((ln(1+(√x)))/(√x))dx


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Question Number 74793 by mathmax by abdo last updated on 30/Nov/19

$p r o v e t h e c o n v e r g e n c e o f \int_{0}^{1} \frac{l n (1 + \sqrt{x})}{\sqrt{x}} d x$
Commented by mathmax by abdo last updated on 06/Dec/19
$I=∫_0 ^1 ((ln(1+(√x)))/(√x))dx changement (√x)=t give x=t^2 ⇒ I =∫_0 ^1 ((ln(1+t))/t)(2t)dt =2 ∫_0 ^1 ln(1+t)dt =_(1+t=u) 2∫_1 ^2 ln(u)du =2[ulnu−u]_1 ^2 =2{2ln(2)−2+1} =4ln(2)−2$
$I = \int_{0}^{1} \frac{l n (1 + \sqrt{x})}{\sqrt{x}} d x c h a n g e m e n t \sqrt{x} = t g i v e x = t^{2} \Rightarrow$ $I = \int_{0}^{1} \frac{l n (1 + t)}{t} (2 t) d t = 2 \int_{0}^{1} l n (1 + t) d t =_{1 + t = u} 2 \int_{1}^{2} l n (u) d u$ $= 2 {[u l n u - u]}_{1}^{2} = 2 {2 l n (2) - 2 + 1} = 4 l n (2) - 2$
	Answered by mind is power last updated on 01/Dec/19

	$u = \sqrt{x} \Rightarrow du = \frac{1}{2 \sqrt{x}} dx$ $\int_{0}^{1} 2 \ln (1 + u) du = {[2 (u + 1) \ln (u + 1) - 2 u]}_{0}^{1} = 4 \ln (2) - 2$
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