Question Number 121117 by benjo_mathlover last updated on 05/Nov/20
$$\:\mathrm{J}=\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{x}−\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{1}}}\:\mathrm{dx}\:? \\ $$
Answered by liberty last updated on 05/Nov/20
$$\Rightarrow\mathrm{2x}^{\mathrm{2}} +\mathrm{x}−\mathrm{1}=\left(\mathrm{2x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)\: \\ $$$$\mathrm{let}\:\sqrt{\mathrm{x}+\mathrm{1}}\:=\:\rho\:\Rightarrow\mathrm{x}=\rho^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{J}=\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\frac{\left(\mathrm{2}\rho^{\mathrm{2}} −\mathrm{3}\right)\rho^{\mathrm{2}} \:\left(\mathrm{2}\rho\:\mathrm{d}\rho\right)}{\rho} \\ $$$$\mathrm{J}=\int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{4}\rho^{\mathrm{4}} −\mathrm{6}\rho^{\mathrm{2}} \right)\:\mathrm{d}\rho\:=\frac{\mathrm{4}}{\mathrm{5}}\left(\mathrm{32}−\mathrm{1}\right)−\mathrm{2}\left(\mathrm{8}−\mathrm{1}\right) \\ $$$$=\:\frac{\mathrm{124}}{\mathrm{5}}−\mathrm{14}=\frac{\mathrm{124}−\mathrm{70}}{\mathrm{5}}=\frac{\mathrm{54}}{\mathrm{5}} \\ $$
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}](https://www.tinkutara.com/question/Q121198.png)
$${J}\:=\int_{\mathrm{0}} ^{\mathrm{3}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}}{dx}\:{we}\:{do}\:{the}\:{changement} \\ $$$$\sqrt{{x}+\mathrm{1}}={t}\:\Rightarrow{x}+\mathrm{1}={t}^{\mathrm{2}\:} \:\Rightarrow{x}={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow \\ $$$${J}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{2}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +{t}^{\mathrm{2}} −\mathrm{1}−\mathrm{1}}{{t}}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\left(\mathrm{2}\left({t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right)+{t}^{\mathrm{2}} −\mathrm{2}\right){dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\left(\mathrm{2}{t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} \right){dt} \\ $$$$=\mathrm{4}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:{t}^{\mathrm{4}} {dt}\:+\mathrm{6}\:\int_{\mathrm{1}} ^{\mathrm{2}} {t}^{\mathrm{2}} {dt} \\ $$$$=\frac{\mathrm{4}}{\mathrm{5}}\left[{t}^{\mathrm{5}} \right]_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{2}\left[{t}^{\mathrm{3}} \right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{5}}\left\{\mathrm{2}^{\mathrm{5}} −\mathrm{1}\right\}\:+\mathrm{2}\left\{\mathrm{2}^{\mathrm{3}} −\mathrm{1}\right\} \\ $$