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Question Number 122430 by benjo_mathlover last updated on 17/Nov/20
∫ (x^3 /( (√(4−x^2 ))+x^2 −4)) dx
$$\:\int\:\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} −\mathrm{4}}\:{dx}\: \\ $$
Answered by liberty last updated on 17/Nov/20
let (√(4−x^2 )) = ω ∧ x^2 −4 =−ω^2 ⇒x dx =−ω dω Q(x)=∫(((4−ω^2 )(−ω dω))/(ω−ω^2 )) Q(x)= ∫ (((ω^2 −4)/(1−ω))) dω Q(x)= ∫ (−ω−1−(3/(1−ω))) dω Q(x)=−(1/2)ω^2 −ω+3ln ∣ω−1∣+c Q(x)=−(1/2)(4−x^2 )−(√(4−x^2 )) + 3ln ∣(√(4−x^2 ))−1∣ + c
$$\:\mathrm{let}\:\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }\:=\:\omega\:\wedge\:\mathrm{x}^{\mathrm{2}} −\mathrm{4}\:=−\omega^{\mathrm{2}} \Rightarrow\mathrm{x}\:\mathrm{dx}\:=−\omega\:\mathrm{d}\omega\: \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)=\int\frac{\left(\mathrm{4}−\omega^{\mathrm{2}} \right)\left(−\omega\:\mathrm{d}\omega\right)}{\omega−\omega^{\mathrm{2}} } \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)=\:\int\:\left(\frac{\omega^{\mathrm{2}} −\mathrm{4}}{\mathrm{1}−\omega}\right)\:\mathrm{d}\omega \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)=\:\int\:\left(−\omega−\mathrm{1}−\frac{\mathrm{3}}{\mathrm{1}−\omega}\right)\:\mathrm{d}\omega \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\omega^{\mathrm{2}} −\omega+\mathrm{3ln}\:\mid\omega−\mathrm{1}\mid+\mathrm{c} \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}−\mathrm{x}^{\mathrm{2}} \right)−\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }\:+\:\mathrm{3ln}\:\mid\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }−\mathrm{1}\mid\:+\:\mathrm{c}\: \\ $$

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