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dx-x-3-x-5-1-3-1-5-




Question Number 127851 by bemath last updated on 02/Jan/21
 ψ = ∫ (dx/(x^3  (((x^5 +1)^3 ))^(1/5) )) ?
$$\:\psi\:=\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \:\sqrt[{\mathrm{5}}]{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{1}\right)^{\mathrm{3}} }}\:?\: \\ $$
Answered by liberty last updated on 02/Jan/21
 ψ = ∫ (dx/(x^3  ((x^(15) (1+x^(−5) )^3 ))^(1/5) ))   ψ = ∫ (dx/(x^6  (((1+x^(−5) )^3 ))^(1/5) )) = ∫ ((x^(−6)  dx)/( (((1+x^(−5) )^3 ))^(1/5) ))   let t = 1+x^(−5)  ⇒dt=−5x^(−6)  dx   ψ = −(1/5)∫ (dt/t^(3/5) ) = −(1/5)∫ t^(−3/5)  dt    ψ =−(1/5). (5/2) t^(2/5)  + C =−((((x^5 +1)^2 ))^(1/5) /(2x^2 ))  +C
$$\:\psi\:=\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \:\sqrt[{\mathrm{5}}]{\mathrm{x}^{\mathrm{15}} \left(\mathrm{1}+\mathrm{x}^{−\mathrm{5}} \right)^{\mathrm{3}} }} \\ $$$$\:\psi\:=\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{6}} \:\sqrt[{\mathrm{5}}]{\left(\mathrm{1}+\mathrm{x}^{−\mathrm{5}} \right)^{\mathrm{3}} }}\:=\:\int\:\frac{\mathrm{x}^{−\mathrm{6}} \:\mathrm{dx}}{\:\sqrt[{\mathrm{5}}]{\left(\mathrm{1}+\mathrm{x}^{−\mathrm{5}} \right)^{\mathrm{3}} }} \\ $$$$\:\mathrm{let}\:\mathrm{t}\:=\:\mathrm{1}+\mathrm{x}^{−\mathrm{5}} \:\Rightarrow\mathrm{dt}=−\mathrm{5x}^{−\mathrm{6}} \:\mathrm{dx} \\ $$$$\:\psi\:=\:−\frac{\mathrm{1}}{\mathrm{5}}\int\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{3}/\mathrm{5}} }\:=\:−\frac{\mathrm{1}}{\mathrm{5}}\int\:\mathrm{t}^{−\mathrm{3}/\mathrm{5}} \:\mathrm{dt}\: \\ $$$$\:\psi\:=−\frac{\mathrm{1}}{\mathrm{5}}.\:\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{t}^{\mathrm{2}/\mathrm{5}} \:+\:\mathrm{C}\:=−\frac{\sqrt[{\mathrm{5}}]{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{2x}^{\mathrm{2}} }\:\:+\mathrm{C} \\ $$

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