J=∫_0 ^3 ((2x^2 +x−1)/( (√(x+1)))) dx ?


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Question Number 121117 by benjo_mathlover last updated on 05/Nov/20

$J = \underset{0}{\int^{3}} \frac{{2 x}^{2} + x - 1}{\sqrt{x + 1}} dx ?$
	Answered by liberty last updated on 05/Nov/20

	$\Rightarrow {2 x}^{2} + x - 1 = (2 x - 1) (x + 1)$ $let \sqrt{x + 1} = ρ \Rightarrow x = ρ^{2} - 1$ $J = \underset{1}{\int^{2}} \frac{(2 ρ^{2} - 3) ρ^{2} (2 ρ d ρ)}{ρ}$ $J = \int_{1}^{2} (4 ρ^{4} - 6 ρ^{2}) d ρ = \frac{4}{5} (32 - 1) - 2 (8 - 1)$ $= \frac{124}{5} - 14 = \frac{124 - 70}{5} = \frac{54}{5}$
	Answered by Bird last updated on 05/Nov/20
	$J =∫_0 ^3 ((2x^2 +x−1)/( (√(x+1))))dx we do the changement (√(x+1))=t ⇒x+1=t^(2 ) ⇒x=t^2 −1 ⇒ J =∫_1 ^2 ((2(t^2 −1)^2 +t^2 −1−1)/t)(2t)dt =2 ∫_1 ^2 (2(t^4 −2t^2 +1)+t^2 −2)dt =2 ∫_1 ^2 (2t^4 −3t^2 )dt =4 ∫_1 ^2 t^4 dt +6 ∫_1 ^2 t^2 dt =(4/5)[t^5 ]_1 ^2 +2[t^3 ]_1 ^2 =(4/5){2^5 −1} +2{2^3 −1}$
	$J = \int_{0}^{3} \frac{2 x^{2} + x - 1}{\sqrt{x + 1}} d x w e d o t h e c h a n g e m e n t$ $\sqrt{x + 1} = t \Rightarrow x + 1 = t^{2} \Rightarrow x = t^{2} - 1 \Rightarrow$ $J = \int_{1}^{2} \frac{2 {(t^{2} - 1)}^{2} + t^{2} - 1 - 1}{t} (2 t) d t$ $= 2 \int_{1}^{2} (2 (t^{4} - 2 t^{2} + 1) + t^{2} - 2) d t$ $= 2 \int_{1}^{2} (2 t^{4} - 3 t^{2}) d t$ $= 4 \int_{1}^{2} t^{4} d t + 6 \int_{1}^{2} t^{2} d t$ $= \frac{4}{5} {[t^{5}]}_{1}^{2} + 2 {[t^{3}]}_{1}^{2}$ $= \frac{4}{5} {2^{5} - 1} + 2 {2^{3} - 1}$
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