∫_4 ^9 ((ℓn (x))/( (√x))) dx ?


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Question Number 121965 by bemath last updated on 13/Nov/20

$\underset{4}{\int^{9}} \frac{ℓ n (x)}{\sqrt{x}} d x ?$
	Answered by bemath last updated on 13/Nov/20
	$∫_4 ^9 ((ℓn(x))/( (√x))) dx = ∫_2 ^3 ((ℓn(u^2 ))/u)(2u du) = ∫_2 ^3 4ℓn(u) du = 4 ( u ℓn (u)−u ) ]_2 ^3 = 4{(3(ℓn(3)−1)−(2(ℓn(2)−1)} = 4(ℓn(((27)/4))−1)$
	$\underset{4}{\int^{9}} \frac{ℓ n (x)}{\sqrt{x}} d x = \underset{2}{\int^{3}} \frac{ℓ n (u^{2})}{u} (2 u d u)$ $= \underset{2}{\int^{3}} 4 ℓ n (u) d u$ ${= 4 (u ℓ n (u) - u)]}_{2}^{3}$ $= 4 {(3 (ℓ n (3) - 1) - (2 (ℓ n (2) - 1)}$ $= 4 (ℓ n (\frac{27}{4}) - 1)$
	Answered by Bird last updated on 13/Nov/20
	$∫_4 ^(9 ) ((lnx)/( (√x)))dx =_((√x)=t) ∫_2 ^(3 ) ((2lnt)/t)(2tdt) =4∫_2 ^3 lnt dt =4[tlnt −t]_2 ^3 =4{3ln3−3−2ln2+2} =4{3ln(3)−2ln2−1}$
	$\int_{4}^{9} \frac{l n x}{\sqrt{x}} d x =_{\sqrt{x} = t} \int_{2}^{3} \frac{2 l n t}{t} (2 t d t)$ $= 4 \int_{2}^{3} l n t d t = 4 {[t l n t - t]}_{2}^{3}$ $= 4 {3 l n 3 - 3 - 2 l n 2 + 2}$ $= 4 {3 l n (3) - 2 l n 2 - 1}$
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