Question Number 140002 by mohammad17 last updated on 03/May/21
Answered by MJS_new last updated on 03/May/21
![∫(x^2 /2)ln ((1+(√(1−x^2 )))/x) dx= [t=((1+(√(1−x^2 )))/x) → dx=−((x^2 (√(1−x^2 )))/(1+(√(1−x^2 ))))dt] =−4∫((t^2 (t^2 −1))/((t^2 +1)^4 ))ln t dt= [by parts] =((4t^3 )/(3(t^2 +1)^3 ))ln t −(4/3)∫(t^2 /((t^2 +1)^3 ))dt= ∫(t^2 /((t^2 +1)^3 ))dt= [Ostrogradski′s Method] =((t(t^2 −1))/(8(t^2 +1)^2 ))+(1/8)∫(dt/(t^2 +1))=((t(t^2 −1))/(8(t^2 +1)^2 ))+(1/8)arctan t =−((t(t^2 −1))/(6(t^2 +1)^2 ))+((4t^3 )/(3(t^2 +1)^3 ))ln t −(1/6)arctan t = =−((x(√(1−x^2 )))/(12))+(x^3 /6)ln ((1+(√(1−x^2 )))/x) −(1/6)arctan ((1+(√(1−x^2 )))/x) +C ⇒ answer is (π/(24))](https://www.tinkutara.com/question/Q140010.png)
$$\int\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:\rightarrow\:{dx}=−\frac{{x}^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dt}\right] \\ $$$$=−\mathrm{4}\int\frac{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }\mathrm{ln}\:{t}\:{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{3}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\mathrm{ln}\:{t}\:−\frac{\mathrm{4}}{\mathrm{3}}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dt}= \\ $$$$ \\ $$$$\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=\frac{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{8}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{8}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{8}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{arctan}\:{t} \\ $$$$ \\ $$$$=−\frac{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{6}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{3}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\mathrm{ln}\:{t}\:−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{arctan}\:{t}\:= \\ $$$$=−\frac{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{12}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{arctan}\:\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\pi}{\mathrm{24}} \\ $$
Commented by mohammad17 last updated on 03/May/21
$${thank}\:{you}\:{sir} \\ $$