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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