Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 64382 by Chi Mes Try last updated on 17/Jul/19

please  help with workings    ∫Ln[(√)(1−x)+(√)(1+x)]dx

$${please}\:\:{help}\:{with}\:{workings} \\ $$$$ \\ $$$$\int{Ln}\left[\sqrt{}\left(\mathrm{1}−{x}\right)+\sqrt{}\left(\mathrm{1}+{x}\right)\right]{dx} \\ $$

Answered by MJS last updated on 17/Jul/19

first step: by parts  u′=1 → u=x  v=ln ((√(1−x))+(√(1+x))) → v′=(1/(2x))−(1/(2x(√(1−x^2 ))))  ∫u′v=uv−∫uv′=xln ((√(1−x))+(√(1+x))) −∫((1/2)−(1/(2(√(1−x^2 )))))dx  second step: standard integrals  −(1/2)∫dx+(1/2)∫(dx/(√(1−x^2 )))=−(x/2)+(1/2)arcsin x    ∫ln ((√(1−x))+(√(1+x))) dx=−(x/2)+xln ((√(1−x))+(√(1+x))) +(1/2)arcsin x +C

$$\mathrm{first}\:\mathrm{step}:\:\mathrm{by}\:\mathrm{parts} \\ $$$${u}'=\mathrm{1}\:\rightarrow\:{u}={x} \\ $$$${v}=\mathrm{ln}\:\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{1}+{x}}\right)\:\rightarrow\:{v}'=\frac{\mathrm{1}}{\mathrm{2}{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\int{u}'{v}={uv}−\int{uv}'={x}\mathrm{ln}\:\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{1}+{x}}\right)\:−\int\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right){dx} \\ $$$$\mathrm{second}\:\mathrm{step}:\:\mathrm{standard}\:\mathrm{integrals} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=−\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcsin}\:{x} \\ $$$$ \\ $$$$\int\mathrm{ln}\:\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{1}+{x}}\right)\:{dx}=−\frac{{x}}{\mathrm{2}}+{x}\mathrm{ln}\:\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{1}+{x}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcsin}\:{x}\:+{C} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com