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Question-125199




Question Number 125199 by joki last updated on 09/Dec/20
Answered by bemath last updated on 09/Dec/20
I= ∫ x (√((√4) x^2 +π^2 )) dx =   I = (1/4)∫ (√(2x^2 +π^2 )) d(2x^2 +π^2 )  I= (1/4)×(2/3) (√((2x^2 +π^2 )^3 )) + c  I= (1/6) (√((2x^2 +π^2 )^3 )) + c
$${I}=\:\int\:{x}\:\sqrt{\sqrt{\mathrm{4}}\:{x}^{\mathrm{2}} +\pi^{\mathrm{2}} }\:{dx}\:=\: \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\pi^{\mathrm{2}} }\:{d}\left(\mathrm{2}{x}^{\mathrm{2}} +\pi^{\mathrm{2}} \right) \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{2}}{\mathrm{3}}\:\sqrt{\left(\mathrm{2}{x}^{\mathrm{2}} +\pi^{\mathrm{2}} \right)^{\mathrm{3}} }\:+\:{c} \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{6}}\:\sqrt{\left(\mathrm{2}{x}^{\mathrm{2}} +\pi^{\mathrm{2}} \right)^{\mathrm{3}} }\:+\:{c}\: \\ $$

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