Question-107656

Question Number 107656 by bemath last updated on 12/Aug/20

Answered by john santu last updated on 12/Aug/20

⋇JS⋇ { ((λ=1+x(√x))),((dx=((2dλ)/(3(√x))))) :} I=∫ (√x) (√(1+x(√x) )) dx I= ∫ (√x) (√λ) (((2dλ)/(3(√x))))=∫(2/3)λ^(1/2) dλ I= (2/3) ((2/3)λ^(3/2) ) +C I= (4/9)(1+x(√x))^(3/2) + C

$$\:\:\:\:\:\divideontimes\mathcal{JS}\divideontimes \\ $$$$\:\begin{cases}{\lambda=\mathrm{1}+{x}\sqrt{{x}}}\\{{dx}=\frac{\mathrm{2}{d}\lambda}{\mathrm{3}\sqrt{{x}}}}\end{cases} \\ $$$${I}=\int\:\sqrt{{x}}\:\sqrt{\mathrm{1}+{x}\sqrt{{x}}\:}\:{dx}\: \\ $$$${I}=\:\int\:\sqrt{{x}}\:\sqrt{\lambda}\:\left(\frac{\mathrm{2}{d}\lambda}{\mathrm{3}\sqrt{{x}}}\right)=\int\frac{\mathrm{2}}{\mathrm{3}}\lambda^{\frac{\mathrm{1}}{\mathrm{2}}} \:{d}\lambda \\ $$$${I}=\:\frac{\mathrm{2}}{\mathrm{3}}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\lambda^{\frac{\mathrm{3}}{\mathrm{2}}} \right)\:+{C}\: \\ $$$${I}=\:\frac{\mathrm{4}}{\mathrm{9}}\left(\mathrm{1}+{x}\sqrt{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\:{C}\: \\ $$

Answered by 1549442205PVT last updated on 12/Aug/20

(√(x−x^2 (√x)))=(√x).(√(1−x(√x))).Hence, Set 1−x(√x)=t^2 ⇒(√(1−x(√x)))=t −((√x)+((√x)/2))dx=2tdt⇒dx=−((4tdt)/(3(√x))) I=−∫((√x).(√(1−x(√x) )) )dx=−∫(( 4t^2 )/3)dt =−((4t^3 )/9)+C=−(4/9)( (√(1−x(√x))) )^3 +C

$$\sqrt{\mathrm{x}−\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}}=\sqrt{\mathrm{x}}.\sqrt{\mathrm{1}−\mathrm{x}\sqrt{\mathrm{x}}}.\mathrm{Hence}, \\ $$$$\mathrm{Set}\:\mathrm{1}−\mathrm{x}\sqrt{\mathrm{x}}=\mathrm{t}^{\mathrm{2}} \Rightarrow\sqrt{\mathrm{1}−\mathrm{x}\sqrt{\mathrm{x}}}=\mathrm{t} \\ $$$$−\left(\sqrt{\mathrm{x}}+\frac{\sqrt{\mathrm{x}}}{\mathrm{2}}\right)\mathrm{dx}=\mathrm{2tdt}\Rightarrow\mathrm{dx}=−\frac{\mathrm{4tdt}}{\mathrm{3}\sqrt{\mathrm{x}}} \\ $$$$\mathrm{I}=−\int\left(\sqrt{\mathrm{x}}.\sqrt{\mathrm{1}−\mathrm{x}\sqrt{\mathrm{x}}\:}\:\right)\mathrm{dx}=−\int\frac{\:\:\mathrm{4t}^{\mathrm{2}} \:}{\mathrm{3}}\mathrm{dt} \\ $$$$=−\frac{\mathrm{4}\boldsymbol{\mathrm{t}}^{\mathrm{3}} }{\mathrm{9}}+\boldsymbol{\mathrm{C}}=−\frac{\mathrm{4}}{\mathrm{9}}\left(\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}\sqrt{\boldsymbol{\mathrm{x}}}}\:\right)^{\mathrm{3}} +\boldsymbol{\mathrm{C}} \\ $$

Commented by bemath last updated on 12/Aug/20

$${typo}\:{sir}\:{it}\:\sqrt{{x}+{x}^{\mathrm{2}} \sqrt{{x}}}\: \\ $$

Answered by Dwaipayan Shikari last updated on 12/Aug/20

∫(√x) ( (√(1+x(√x))))dx 1+x(√x) =t , (3/2)(√x) =(dt/dx) (2/3)∫(3/2)(√x) ((√(1+x(√x))))dx (2/3)∫(√t) dt (4/9)t^(3/2) +C=(4/9)(1+x(√x))^(3/2) +C

$$\int\sqrt{{x}}\:\left(\:\sqrt{\mathrm{1}+{x}\sqrt{{x}}}\right){dx}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}+{x}\sqrt{{x}}\:={t}\:\:,\:\frac{\mathrm{3}}{\mathrm{2}}\sqrt{{x}}\:=\frac{{dt}}{{dx}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\mathrm{3}}{\mathrm{2}}\sqrt{{x}}\:\left(\sqrt{\mathrm{1}+{x}\sqrt{{x}}}\right){dx} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int\sqrt{{t}}\:{dt} \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}{t}^{\frac{\mathrm{3}}{\mathrm{2}}} +{C}=\frac{\mathrm{4}}{\mathrm{9}}\left(\mathrm{1}+{x}\sqrt{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +{C} \\ $$

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