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Question Number 51421 by Tawa1 last updated on 26/Dec/18

$$\int \frac{{\tan }^{-1}\mathrm{x}}{{\mathrm{x}}^2}\mathrm{dx}$$

Commented by Tawa1 last updated on 27/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by maxmathsup by imad last updated on 26/Dec/18

$$letdeterminef\left(t\right)=\int \frac{arctan\left(tx\right)}{x^2}dxwitht>0byparts$$$$f\left(t\right)=-\frac{1}{x}arctan\left(tx\right)-\int -\frac{1}{x}\frac{t}{1+t^2{x}^2}dx$$$$=-\frac{arctan\left(tx\right)}{x}+t\int \frac{dx}{x(1+t^2{x}^2)}but$$$$\int \frac{dx}{x(1+t^2{x}^2)}{=}_{tx=u}\int \frac{du}{t\frac{u}{t}(1+u^2)}=\int \frac{du}{u(1+u^2)}$$$$=\int (\frac{1}{u}-\frac{u}{1+u^2})=ln\mid u\mid -\frac{1}{2}ln(1+u^2)=ln\left(\frac{\mid u\mid }{\sqrt{1+u^2}}\right)+c$$$$=ln\left(\frac{\mid tx\mid }{\sqrt{1+t^2{x}^2}}\right)\Rightarrow f\left(t\right)=-\frac{arctan\left(tx\right)}{x}+tln\left(\frac{\mid tx\mid }{\sqrt{1+t^2{x}^2}}\right)+cand$$$$\int \frac{arctan\left(x\right)}{x^2}dx=f\left(1\right)=-\frac{arctanx}{x}+ln\left(\frac{\mid x\mid }{\sqrt{1+x^2}}\right)+C.$$

Commented by Abdo msup. last updated on 30/Dec/18

$$youarewelcomesir.$$

Answered by ajfour last updated on 26/Dec/18

$$I=-\frac{{\tan }^{-1}x}{x}+\int \frac{dx}{x(1+x^2)}$$$$Now\frac{1}{x(1+x^2)}=\frac{A}{x}+\frac{Bx+C}{1+x^2}$$$$\Rightarrow 1=A+{Ax}^2+{Bx}^2+Cx$$$$\Rightarrow A=1,C=0,B=-1$$$$I=-\frac{{\tan }^{-1}x}{x}+\int \frac{dx}{x}-\int \frac{x}{1+x^2}dx$$$$I=-\frac{{\tan }^{-1}x}{x}+\ln \mid x\mid -\frac{1}{2}\ln \mid 1+x^2\mid +c.$$

Commented by Tawa1 last updated on 26/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by Smail last updated on 26/Dec/18

$$byparts$$$$u={tan}^{-1}x\Rightarrow u^{\prime }=\frac{1}{1+x^2}$$$$v^{\prime }=\frac{1}{x^2}\Rightarrow v=\frac{-1}{x}$$$$A=\int \frac{{tan}^{-1}x}{x^2}dx=\frac{-{tan}^{-1}x}{x}+\int \frac{dx}{x(1+x^2)}+c_1$$$$\frac{1}{x(1+x^2)}=\frac{a}{x}+\frac{bx+c}{x^2+1}$$$$witha=1,b=-1,c=0$$$$A=\frac{-{tan}^{-1}x}{x}+\int (\frac{1}{x}-\frac{x}{x^2+1})dx+c_1$$$$=-\frac{{tan}^{-1}x}{x}+ln\mid x\mid -\frac{1}{2}ln(x^2+1)+C$$$$\int \frac{{tan}^{-1}x}{x^2}dx=-\frac{{tan}^{-1}x}{x}+ln\mid \frac{x}{\sqrt{x^2+1}}\mid +C$$

Commented by Tawa1 last updated on 26/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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