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Question Number 158742 by cortano last updated on 08/Nov/21

Commented by HongKing last updated on 08/Nov/21

= (2/5) (x^6 + x^4 + (1/x^4 ))^(5/4) + C

$$=\:\frac{\mathrm{2}}{\mathrm{5}}\:\left(\mathrm{x}^{\mathrm{6}} \:+\:\mathrm{x}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right)^{\frac{\mathrm{5}}{\mathrm{4}}} +\:\mathbb{C} \\ $$

Answered by tounghoungko last updated on 08/Nov/21

∫ ((x^5 (3x^5 +2x^3 −2x^(−5) )x ((x^6 +x^4 +x^(−4) ))^(1/4) )/x^6 ) dx =∫ (3x^5 +2x^3 −2x^(−5) ) ((x^6 +x^4 +x^(−4) ))^(1/4) dx let λ=x^6 +x^4 +x^(−4) ⇒dλ=6x^5 +4x^3 −4x^(−5) dx I=(1/2)∫ λ^(1/4) dλ= (2/5)λ^(5/4) +c =(2/5)(((x^6 +x^4 +x^(−4) )^5 ))^(1/4) + c =(2/5) ((((x^(10) +x^8 +1)^5 )/x^(20) ))^(1/4) +c =((2 (((x^(10) +x^8 +1)^5 ))^(1/4) )/(5x^( 5) )) + c

$$\int\:\frac{{x}^{\mathrm{5}} \left(\mathrm{3}{x}^{\mathrm{5}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{−\mathrm{5}} \right){x}\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{−\mathrm{4}} }}{{x}^{\mathrm{6}} }\:{dx} \\ $$$$=\int\:\left(\mathrm{3}{x}^{\mathrm{5}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{−\mathrm{5}} \right)\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{−\mathrm{4}} }\:{dx} \\ $$$${let}\:\lambda={x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{−\mathrm{4}} \:\Rightarrow{d}\lambda=\mathrm{6}{x}^{\mathrm{5}} +\mathrm{4}{x}^{\mathrm{3}} −\mathrm{4}{x}^{−\mathrm{5}} \:{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\:\lambda^{\mathrm{1}/\mathrm{4}} \:{d}\lambda=\:\frac{\mathrm{2}}{\mathrm{5}}\lambda^{\mathrm{5}/\mathrm{4}} \:+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\sqrt[{\mathrm{4}}]{\left({x}^{\mathrm{6}} +{x}^{\mathrm{4}} +{x}^{−\mathrm{4}} \right)^{\mathrm{5}} }\:+\:{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\:\sqrt[{\mathrm{4}}]{\frac{\left({x}^{\mathrm{10}} +{x}^{\mathrm{8}} +\mathrm{1}\right)^{\mathrm{5}} }{{x}^{\mathrm{20}} }}\:+{c} \\ $$$$=\frac{\mathrm{2}\:\sqrt[{\mathrm{4}}]{\left({x}^{\mathrm{10}} +{x}^{\mathrm{8}} +\mathrm{1}\right)^{\mathrm{5}} }}{\mathrm{5}{x}^{\:\mathrm{5}} }\:+\:{c}\: \\ $$

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