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Question Number 41135 by math khazana by abdo last updated on 02/Aug/18

$$findf\left(x\right)={\int }_0^{1}arctan\left(xt\right)dtxfromR$$

Answered by math khazana by abdo last updated on 03/Aug/18

$$wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^1\frac{t}{1+x^2{t}^2}dt$$$${=}_{xt=u}{\int }_0^x\frac{u}{x(1+u^2)}\frac{du}{x}=\frac{1}{x^2}{\int }_0^x\frac{udu}{1+u^2}$$$$=\frac{1}{2x^2}{\left[ln(1+u^2)\right]}_0^x=\frac{ln(1+x^2)}{2x^2}\Rightarrow$$$$f\left(x\right)={\int }_0^x\frac{ln(1+t^2)}{2t^2}dt+c$$$$c=f\left(0\right)=0\Rightarrow f\left(x\right)={\int }_0^x\frac{ln(1+t^2)}{2t^2}dt$$$$byparts2f\left(x\right)={[-\frac{1}{t}ln(1+t^2)]}_0^x+{\int }_0^x\frac{1}{t}\frac{2t}{1+t^2}dt$$$$=-\frac{1}{x}ln(1+x^2)+2arctanx\Rightarrow$$$$f\left(x\right)=arctan\left(x\right)-\frac{1}{2x}ln(1+x^2).$$

Commented by math khazana by abdo last updated on 03/Aug/18

$$letprove{lim}_{t\to 0}\frac{ln(1+t^2)}{t}=0wehave$$$$ln(1+t^2)\sim t^2\Rightarrow \frac{ln(1+t^2)}{t}\sim t(t\to 0)resultisproved$$

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