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Question Number 59168 by maxmathsup by imad last updated on 05/May/19

calculate ∫_(−1) ^1 ((x^2 +3)/((√(1+x)) +(√(1−x)))) dx

$${calculate}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}} \:+\mathrm{3}}{\sqrt{\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}}\:{dx} \\ $$

Answered by ajfour last updated on 05/May/19

(I/2)=∫_0 ^( 1) ((x^2 +3)/((√(1+x))+(√(1−x))))dx =∫_0 ^( 1) (((x^2 +3)((√(1+x))−(√(1−x))))/(2x))dx I=∫_0 ^( 1) xdx−3∫_0 ^( (π/2)) (((√2)cos θ−(√2)sin θ)/((cos^2 θ−sin^2 θ)))(−4sin θcos θdθ) =(1/2)+12(√2)∫_0 ^( (π/2)) ((sin θcos θ)/(cos θ+sin θ))dθ = (1/2)+24(√2)∫_0 ^( π/4) ((sin θdθ)/(1+tan θ)) ......

$$\frac{\mathrm{I}}{\mathrm{2}}=\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \frac{\mathrm{x}^{\mathrm{2}} +\mathrm{3}}{\sqrt{\mathrm{1}+\mathrm{x}}+\sqrt{\mathrm{1}−\mathrm{x}}}\mathrm{dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right)\left(\sqrt{\mathrm{1}+\mathrm{x}}−\sqrt{\mathrm{1}−\mathrm{x}}\right)}{\mathrm{2x}}\mathrm{dx} \\ $$$$\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \mathrm{xdx}−\mathrm{3}\int_{\mathrm{0}} ^{\:\:\frac{\pi}{\mathrm{2}}} \frac{\sqrt{\mathrm{2}}\mathrm{cos}\:\theta−\sqrt{\mathrm{2}}\mathrm{sin}\:\theta}{\left(\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\right)}\left(−\mathrm{4sin}\:\theta\mathrm{cos}\:\theta\mathrm{d}\theta\right) \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{12}\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{cos}\:\theta+\mathrm{sin}\:\theta}\mathrm{d}\theta \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{24}\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{4}} \frac{\mathrm{sin}\:\theta\mathrm{d}\theta}{\mathrm{1}+\mathrm{tan}\:\theta} \\ $$$$...... \\ $$

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