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Question Number 212307 by MrGaster last updated on 09/Oct/24

$$$$$$I={\int }_0^{\mathrm{\infty }}\frac{{(x-\arctan x)}^2}{x^4}dx.$$

Answered by Ghisom last updated on 09/Oct/24

$$\int \frac{{(x-\arctan x)}^2}{x^4}dx=A+B+C$$$$A=\int \frac{dx}{x^2}=-\frac{1}{x}$$$$B=-2\int \frac{\arctan x}{x^3}dx=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=\frac{1}{x}+\frac{x^2+1}{x^2}\arctan x$$$$A+B=\frac{x^2+1}{x^2}\arctan x$$$$C=\int \frac{{\arctan }^{2}x}{x^4}dx=$$$$[t=\arctan x]$$$$=\int \frac{t^2{\cos }^{2}t}{{\sin }^{4}t}dt=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=D+E$$$$D=-\frac{t^2}{{3\tan }^{3}t}=-\frac{{\arctan }^{2}x}{3x^3}$$$$E=\frac{2}{3}\int \frac{t}{{\tan }^{3}t}dt=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=F+G$$$$F=-\frac{t}{{3\tan }^{2}t}-\frac{2t}{3}\ln \sin t$$$$F=\frac{\arctan x}{3}(\ln (x^2+1)-2\ln \mid x\mid -\frac{1}{x^2})$$$$G=H+J$$$$H=\frac{1}{3}\int \frac{dt}{{\tan }^{2}t}=-\frac{1}{3}(t+\frac{1}{\tan t})$$$$H=-\frac{1}{3}(\frac{1}{x}+\arctan x)$$$$\mathrm{all}\mathrm{above}x\in [0,\mathrm{\infty })$$$$J=\frac{2}{3}\underset{0}{\stackrel{\pi /2}{\int }}\ln \sin tdt=-\frac{\pi \ln 2}{3}$$$$[\mathrm{this}\mathrm{is}\mathrm{easy};-)]$$$$\mathrm{I}\mathrm{get}$$$$I=\frac{\pi }{3}(1-\ln 2)$$

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