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Question Number 60691 by maxmathsup by imad last updated on 24/May/19

$$calculatef\left(a\right)=\int (1-\frac{a}{x^2})arctan(x+\frac{a}{x})dxwithareal.$$

Commented by maxmathsup by imad last updated on 27/May/19

$$byparts{u}^{{}^{\prime }}=1-\frac{a}{x^2}andv=arctan(x+\frac{a}{x})\Rightarrow$$$$f\left(a\right)=(x+\frac{a}{x})arctan(x+\frac{a}{x})-\int (x+\frac{a}{x})\frac{1-\frac{a}{x^2}}{1+{(x+\frac{a}{x})}^2}dx$$$$=(x+\frac{a}{x})arctan(x+\frac{a}{x})-\int (x+\frac{a}{x})\frac{x^2-a}{x^2+{(x^2+a^2)}^2}dxbut$$$$\int (x+\frac{a}{x})\frac{x^2-a}{x^2+{(x^2+a^2)}^2}=\int \frac{x^4-a^2}{x(x^2+x^4+2x^2{a}^2+a^4)}dx$$$$=\int \frac{x^4-a^2}{x(x^4+(2a^2+1)x^2+a^4)}dxletF\left(x\right)=\frac{x^4-a^2}{x(x^4+(2a^2+1)x^2+a^4)}$$$$polesofF?$$$$rootsof{x}^4+(2a^2+1)x^2+a^4=0\Rightarrow t^2+(2a^2+1)t+a^4=0(t=x^2)$$$$\mathrm{\Delta }={(2a^2+1)}^2-4a^4=4a^4+4a^2+1-4a^4=4a^2+1\Rightarrow$$$$t_1=\frac{-2a^2-1+\sqrt{4a^2+1}}{2}and{t}_2=\frac{-2a^2-1-\sqrt{4a^2+1}}{2}$$$$F\left(x\right)=\frac{x^4-a^2}{x(x^2-t_1)(x^2-t_2)}=\frac{x^4-a^2}{x(x^2+\frac{2a^2+1-\sqrt{4a^2+1}}{2})(x^2+\frac{2a^2+1+\sqrt{4a^2+}1}{2})}$$$$thedecompositionofF\left(x\right)isatform$$$$F\left(x\right)=\frac{a}{x}+\frac{bx+c}{x^2+\frac{2a^2+1-\sqrt{4a^2+1}}{2}}+\frac{dx+e}{x^2+\frac{2a^2+1+\sqrt{4a^2+1}}{2}}$$$$a={lim}_{x\to a}xF\left(x\right)=\frac{a^4-a^2}{(a^2-t_1)(a^2-t_2)}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=1=a+b+d\Rightarrow b+d=-a....becontinued....$$

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