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Question Number 123144 by aurpeyz last updated on 23/Nov/20

∫((x−1)/( (√(6x−5−x^2 ))))dx

$$\int\frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{6}{x}−\mathrm{5}−{x}^{\mathrm{2}} }}{dx} \\ $$

Answered by benjo_mathlover last updated on 23/Nov/20

6x−5−x^2 = −5−(x^2 −6x) = −5−(x^2 −6x+9)+9 = 4−(x−3)^2 ∅(x)=∫ ((x−1)/( (√(4−(x−3)^2 )))) dx let x−3 = 2sin r ; dx = 2cos r dr x−1 = 2sin r+2 ∅(x)=∫ ((2(sin r+1)(2cos r dr))/( (√(4−4sin^2 r)))) ∅(x)= 2∫(sin r+1)dr ∅(x)=−2cos r+2r + c ∅(x)=−2 (((√(4−(x−3)^2 ))/2))+2sin^(−1) (((x−3)/2))+c ∅(x)=−(√(6x−5−x^2 )) + 2sin^(−1) (((x−3)/2)) + c

$$\mathrm{6}{x}−\mathrm{5}−{x}^{\mathrm{2}} \:=\:−\mathrm{5}−\left({x}^{\mathrm{2}} −\mathrm{6}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\mathrm{5}−\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right)+\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}−\left({x}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\emptyset\left({x}\right)=\int\:\frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{4}−\left({x}−\mathrm{3}\right)^{\mathrm{2}} }}\:{dx}\: \\ $$$$\:{let}\:{x}−\mathrm{3}\:=\:\mathrm{2sin}\:{r}\:;\:{dx}\:=\:\mathrm{2cos}\:{r}\:{dr} \\ $$$${x}−\mathrm{1}\:=\:\mathrm{2sin}\:{r}+\mathrm{2}\: \\ $$$$\emptyset\left({x}\right)=\int\:\frac{\mathrm{2}\left(\mathrm{sin}\:{r}+\mathrm{1}\right)\left(\mathrm{2cos}\:{r}\:{dr}\right)}{\:\sqrt{\mathrm{4}−\mathrm{4sin}\:^{\mathrm{2}} {r}}} \\ $$$$\emptyset\left({x}\right)=\:\mathrm{2}\int\left(\mathrm{sin}\:{r}+\mathrm{1}\right){dr} \\ $$$$\emptyset\left({x}\right)=−\mathrm{2cos}\:{r}+\mathrm{2}{r}\:+\:{c}\: \\ $$$$\emptyset\left({x}\right)=−\mathrm{2}\:\left(\frac{\sqrt{\mathrm{4}−\left({x}−\mathrm{3}\right)^{\mathrm{2}} }}{\mathrm{2}}\right)+\mathrm{2sin}^{−\mathrm{1}} \left(\frac{{x}−\mathrm{3}}{\mathrm{2}}\right)+{c} \\ $$$$\emptyset\left({x}\right)=−\sqrt{\mathrm{6}{x}−\mathrm{5}−{x}^{\mathrm{2}} }\:+\:\mathrm{2sin}^{−\mathrm{1}} \left(\frac{{x}−\mathrm{3}}{\mathrm{2}}\right)\:+\:{c} \\ $$

Answered by Dwaipayan Shikari last updated on 23/Nov/20

∫((x−3)/( (√(6x−5−x^2 ))))+(2/( (√(−(x^2 −6x+9)+4))))dx =−(1/2)∫((6−2x)/( (√(6x−5−x^2 ))))+2∫(1/( (√(4−(x−3)^2 ))))dx =−(√(6x−5−x^2 )) +2sin^(−1) ((x−3)/2)+C

$$\int\frac{{x}−\mathrm{3}}{\:\sqrt{\mathrm{6}{x}−\mathrm{5}−{x}^{\mathrm{2}} }}+\frac{\mathrm{2}}{\:\sqrt{−\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right)+\mathrm{4}}}{dx}\:\:\:\:\:\:\:\: \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{6}−\mathrm{2}{x}}{\:\sqrt{\mathrm{6}{x}−\mathrm{5}−{x}^{\mathrm{2}} }}+\mathrm{2}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}−\left({x}−\mathrm{3}\right)^{\mathrm{2}} }}{dx}\:\:\: \\ $$$$=−\sqrt{\mathrm{6}{x}−\mathrm{5}−{x}^{\mathrm{2}} }\:+\mathrm{2}{sin}^{−\mathrm{1}} \frac{{x}−\mathrm{3}}{\mathrm{2}}+{C} \\ $$

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