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Question Number 33328 by prof Abdo imad last updated on 14/Apr/18

$$find{\int }_{\frac{\pi }{4}}^{\frac{4}{\pi }}(1+\frac{1}{x^2})arctanxdx$$

Commented by math khazana by abdo last updated on 19/Apr/18

$$letputI={\int }_{\frac{\pi }{4}}^{\frac{4}{\pi }}(1+\frac{1}{x^2})arctanxdx.letintegrateby$$$$parts{u}^{{}^{\prime }}=1+\frac{1}{x^2}andv=arctanx$$$$I={\left[(1-\frac{1}{x})arctanx\right]}_{\frac{\pi }{4}}^{\frac{4}{\pi }}-{\int }_{\frac{\pi }{4}}^{\frac{4}{\pi }}(1-\frac{1}{x})\frac{dx}{1+x^2}$$$$=(1-\frac{\pi }{4})arctan\left(\frac{4}{\pi }\right)-(1-\frac{4}{\pi })-{\int }_{\frac{\pi }{4}}^{\frac{4}{\pi }}\frac{dx}{1+x^2}$$$$+{\int }_{\frac{\pi }{4}}^{\frac{4}{\pi }}\frac{dx}{x(1+x^2)}but$$$${\int }_{\frac{\pi }{4}}^{\frac{4}{\pi }}\frac{dx}{1+x^2}=arctan\left(\frac{4}{\pi }\right)-arctan\left(\frac{\pi }{4}\right)$$$$=\frac{\pi }{2}-1-1=\frac{\pi }{2}-2let?decompose$$$$F\left(x\right)=\frac{1}{x(1+x^2)}=\frac{a}{x}+\frac{bx+c}{1+x^2}$$$$a={lim}_{x\to 0}xF\left(x\right)=1$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+b\Rightarrow b=-a=-1$$$$F\left(x\right)=\frac{1}{x}+\frac{-x+c}{1+x^2}welooktbatc=0\Rightarrow$$$$F\left(x\right)=\frac{1}{x}-\frac{x}{1+x^2}\Rightarrow$$$${\int }_{\frac{\pi }{4}}^{\frac{4}{\pi }}\frac{dx}{x(1+x^2)}={\int }_{\frac{\pi }{4}}^{\frac{4}{\pi }}(\frac{1}{x}-\frac{x}{1+x^2})dx$$$$={[ln\left(x\right)-\frac{1}{2}ln(1+x^2)]}_{\frac{\pi }{4}}^{\frac{4}{\pi }}={\left[ln\left(\frac{x}{\sqrt{1+x^2}}\right)\right]}_{\frac{\pi }{4}}^{\frac{4}{\pi }}$$$$=ln\left(\frac{4}{\pi \sqrt{1+\frac{16}{{\pi }^2}}}\right)-ln\left(\frac{\pi }{4\sqrt{1+\frac{{\pi }^2}{16}}}\right)$$$$I=(1-\frac{\pi }{4})(\frac{\pi }{2}-1)+1+\frac{4}{\pi }-\frac{\pi }{2}$$$$+ln\left(\frac{4}{\pi \sqrt{1+\frac{16}{{\pi }^2}}}\right)-ln\left(\frac{\pi }{4\sqrt{1+\frac{{\pi }^2}{16}}}\right).$$

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