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Question Number 190746 by leicianocosta last updated on 10/Apr/23

	Answered by Mathspace last updated on 12/Apr/23
	$a)∫_1 ^4 x(√(1+x^2 ))dx=(1/2)∫_1 ^4 (2x)(x^2 +1)^(1/2) dx [(1/3)(x^2 +1)^(3/2) ]_1 ^4 =(1/3){5^(3/2) −2^(3/2) }$
	$a) \int_{1}^{4} x \sqrt{1 + x^{2}} d x = \frac{1}{2} \int_{1}^{4} (2 x) {(x^{2} + 1)}^{\frac{1}{2}} d x$ ${[\frac{1}{3} {(x^{2} + 1)}^{\frac{3}{2}}]}_{1}^{4} = \frac{1}{3} {5^{\frac{3}{2}} - 2^{\frac{3}{2}}}$
	Answered by Mathspace last updated on 12/Apr/23
	$b) ∫_2 ^3 (x/( (√(x−1))))dx we do the changement (√(x−1))=t ⇒x−1=t^2 and ∫_2 ^3 (x/( (√(x−1))))dx=∫_1 ^(√2) ((1+t^2 )/t)(2t)dt =2∫_1 ^(√2) (1+t^2 )dt =2[t+(t^3 /3)]_1 ^(√2) =2{(√2)+((((√2))^3 )/3)−1−(1/3)}$
	$b) \int_{2}^{3} \frac{x}{\sqrt{x - 1}} d x w e d o t h e c h a n g e m e n t$ $\sqrt{x - 1} = t \Rightarrow x - 1 = t^{2} a n d$ $\int_{2}^{3} \frac{x}{\sqrt{x - 1}} d x = \int_{1}^{\sqrt{2}} \frac{1 + t^{2}}{t} (2 t) d t$ $= 2 \int_{1}^{\sqrt{2}} (1 + t^{2}) d t$ $= 2 {[t + \frac{t^{3}}{3}]}_{1}^{\sqrt{2}}$ $= 2 {\sqrt{2} + \frac{{(\sqrt{2})}^{3}}{3} - 1 - \frac{1}{3}}$
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