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x-2-x-x-3-dx-




Question Number 144234 by mathdanisur last updated on 23/Jun/21
∫ ((√(x^2  - x))/x^3 ) dx = ?
$$\int\:\frac{\sqrt{{x}^{\mathrm{2}} \:-\:{x}}}{{x}^{\mathrm{3}} }\:{dx}\:=\:? \\ $$
Answered by liberty last updated on 23/Jun/21
∫ ((√(1−x^(−1) ))/x^2 ) dx = ∫ x^(−2)  (√(1−x^(−1) )) dx  let u^2  = 1−x^(−1)  →2u du = x^(−2)  dx  then ∫ (√u^2 ) (2u du)=∫2u^2  du  = (2/3)u^3  + c = (2/3)(√((1−x^(−1) )^3 )) +c   = ((2(√((x−1)^3 )))/(3x(√x))) + c   =((2(x−1)(√(x−1)))/(3x(√x))) + c
$$\int\:\frac{\sqrt{\mathrm{1}−\mathrm{x}^{−\mathrm{1}} }}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:\int\:\mathrm{x}^{−\mathrm{2}} \:\sqrt{\mathrm{1}−\mathrm{x}^{−\mathrm{1}} }\:\mathrm{dx} \\ $$$$\mathrm{let}\:\mathrm{u}^{\mathrm{2}} \:=\:\mathrm{1}−\mathrm{x}^{−\mathrm{1}} \:\rightarrow\mathrm{2u}\:\mathrm{du}\:=\:\mathrm{x}^{−\mathrm{2}} \:\mathrm{dx} \\ $$$$\mathrm{then}\:\int\:\sqrt{\mathrm{u}^{\mathrm{2}} }\:\left(\mathrm{2u}\:\mathrm{du}\right)=\int\mathrm{2u}^{\mathrm{2}} \:\mathrm{du} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{u}^{\mathrm{3}} \:+\:\mathrm{c}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\left(\mathrm{1}−\mathrm{x}^{−\mathrm{1}} \right)^{\mathrm{3}} }\:+\mathrm{c}\: \\ $$$$=\:\frac{\mathrm{2}\sqrt{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} }}{\mathrm{3x}\sqrt{\mathrm{x}}}\:+\:\mathrm{c}\: \\ $$$$=\frac{\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)\sqrt{\mathrm{x}−\mathrm{1}}}{\mathrm{3x}\sqrt{\mathrm{x}}}\:+\:\mathrm{c} \\ $$
Commented by mathdanisur last updated on 23/Jun/21
Thank you Sir
$${Thank}\:{you}\:{Sir} \\ $$

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