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Question Number 140002 by mohammad17 last updated on 03/May/21

Answered by MJS_new last updated on 03/May/21

$$\int \frac{x^2}{2}\ln \frac{1+\sqrt{1-x^2}}{x}dx=$$$$[t=\frac{1+\sqrt{1-x^2}}{x}\to dx=-\frac{x^2\sqrt{1-x^2}}{1+\sqrt{1-x^2}}dt]$$$$=-4\int \frac{t^2(t^2-1)}{{(t^2+1)}^4}\ln tdt=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=\frac{4t^3}{3{(t^2+1)}^3}\ln t-\frac{4}{3}\int \frac{t^2}{{(t^2+1)}^3}dt=$$$$$$$$\int \frac{t^2}{{(t^2+1)}^3}dt=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{t(t^2-1)}{8{(t^2+1)}^2}+\frac{1}{8}\int \frac{dt}{t^2+1}=\frac{t(t^2-1)}{8{(t^2+1)}^2}+\frac{1}{8}\arctan t$$$$$$$$=-\frac{t(t^2-1)}{6{(t^2+1)}^2}+\frac{4t^3}{3{(t^2+1)}^3}\ln t-\frac{1}{6}\arctan t=$$$$=-\frac{x\sqrt{1-x^2}}{12}+\frac{x^3}{6}\ln \frac{1+\sqrt{1-x^2}}{x}-\frac{1}{6}\arctan \frac{1+\sqrt{1-x^2}}{x}+C$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{\pi }{24}$$

Commented by mohammad17 last updated on 03/May/21

$$thankyousir$$

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