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Question Number 137477 by Sandeep11 last updated on 03/Apr/21

Commented by MJS_new last updated on 03/Apr/21

well just derivate the answers and find the right one?!

$$\mathrm{well}\:\mathrm{just}\:\mathrm{derivate}\:\mathrm{the}\:\mathrm{answers}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{right}\:\mathrm{one}?! \\ $$

Answered by soumyasaha last updated on 03/Apr/21

= ∫ ((x^2 (1−(1/x^2 )))/(x.x(√((x+(1/x)+α)(x+(1/x)+β))))) dx = ∫ (((1−(1/x^2 )))/( (√((x+(1/x)+α)(x+(1/x)+β))))) dx Let, x + (1/x) = t ⇒ (1−(1/x^2 ))dx = dt ∴ I = ∫ (dt/( (√((t+α)(t+β))))) Let, (√(t+α)) = z ⇒ t+α = z^2 ⇒ dt = 2zdz ∴ I = ∫ ((2zdz)/( (√(z^2 (z^2 +β−α))))) = 2.∫ (dz/( (√(z^2 +(β−α))))) = 2ln ∣ z + (√(z^2 +β−α)) ∣ + C = 2 ln ∣ (√(x + (1/x) + α)) + (√(x+(1/x) + β)) ∣ + C = 2 ln ∣ (((√(x^2 +αx+1)) + (√(x^2 +βx+1)))/( (√x))) ∣ + C

$$\:\:\:\:=\:\int\:\frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{x}.\mathrm{x}\sqrt{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}+\alpha\right)\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}+\beta\right)}}\:\mathrm{dx} \\ $$$$\:\:\:\:=\:\int\:\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)}{\:\sqrt{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}+\alpha\right)\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}+\beta\right)}}\:\mathrm{dx} \\ $$$$\:\:\mathrm{Let},\:\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{t}\:\:\Rightarrow\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\mathrm{dx}\:=\:\mathrm{dt} \\ $$$$\:\therefore\:\mathrm{I}\:=\:\int\:\frac{\mathrm{dt}}{\:\sqrt{\left(\mathrm{t}+\alpha\right)\left(\mathrm{t}+\beta\right)}} \\ $$$$\:\:\:\:\mathrm{Let},\:\:\sqrt{\mathrm{t}+\alpha}\:=\:\mathrm{z}\:\:\Rightarrow\:\mathrm{t}+\alpha\:=\:\mathrm{z}^{\mathrm{2}} \:\Rightarrow\:\mathrm{dt}\:=\:\mathrm{2zdz} \\ $$$$\:\therefore\:\mathrm{I}\:=\:\int\:\frac{\mathrm{2zdz}}{\:\sqrt{\mathrm{z}^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} +\beta−\alpha\right)}} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{2}.\int\:\frac{\mathrm{dz}}{\:\sqrt{\mathrm{z}^{\mathrm{2}} +\left(\beta−\alpha\right)}} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{2ln}\:\mid\:\mathrm{z}\:+\:\sqrt{\mathrm{z}^{\mathrm{2}} +\beta−\alpha}\:\mid\:+\:\mathrm{C} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{2}\:\mathrm{ln}\:\mid\:\sqrt{\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\:+\:\alpha}\:+\:\sqrt{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\:+\:\beta}\:\mid\:+\:\mathrm{C} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{2}\:\mathrm{ln}\:\mid\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} +\alpha\mathrm{x}+\mathrm{1}}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} +\beta\mathrm{x}+\mathrm{1}}}{\:\sqrt{\mathrm{x}}}\:\mid\:+\:\mathrm{C} \\ $$

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