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Question Number 218163 by hardmath last updated on 31/Mar/25

Find: ∫ (x/( (√(48 − 2x − x^2 )))) dx = ?

$$\mathrm{Find}:\:\:\:\int\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{48}\:−\:\mathrm{2x}\:−\:\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx}\:=\:? \\ $$

Answered by efronzo1 last updated on 31/Mar/25

I= ∫ (x/( (√(49−(x+1)^2 )))) dx x+1 = 7sin u dx = 7cos u du I = ∫ ((7sin u−1)/( 7(√(1−sin^2 u)))) . 7cos u du = ∫ (7sin u−1) du = −7cos u − u + c = −7(((√(48−2x−x^2 ))/7))−arcsin (((x+1)/7))+c = −(√(48−2x−x^2 ))−arcsin (((x+1)/7))+c

$$\:\mathrm{I}=\:\int\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{49}−\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }}\:\mathrm{dx} \\ $$$$\:\mathrm{x}+\mathrm{1}\:=\:\mathrm{7sin}\:\:\mathrm{u}\: \\ $$$$\:\mathrm{dx}\:=\:\mathrm{7cos}\:\mathrm{u}\:\mathrm{du} \\ $$$$\:\mathrm{I}\:=\:\int\:\frac{\mathrm{7sin}\:\mathrm{u}−\mathrm{1}}{\:\mathrm{7}\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{u}}}\:.\:\mathrm{7cos}\:\mathrm{u}\:\mathrm{du} \\ $$$$\:\:\:=\:\int\:\left(\mathrm{7sin}\:\mathrm{u}−\mathrm{1}\right)\:\mathrm{du} \\ $$$$\:\:=\:−\mathrm{7cos}\:\mathrm{u}\:−\:\mathrm{u}\:+\:\mathrm{c} \\ $$$$\:=\:−\mathrm{7}\left(\frac{\sqrt{\mathrm{48}−\mathrm{2x}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{7}}\right)−\mathrm{arcsin}\:\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{7}}\right)+\mathrm{c} \\ $$$$\:=\:−\sqrt{\mathrm{48}−\mathrm{2x}−\mathrm{x}^{\mathrm{2}} }−\mathrm{arcsin}\:\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{7}}\right)+\mathrm{c}\: \\ $$

Commented by hardmath last updated on 31/Mar/25

thank you dear ser

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{ser} \\ $$

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