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Question Number 28200 by abdo imad last updated on 21/Jan/18

$$letgiveI={\int }_0^1\frac{ln(1+x)}{1+x^2}dxandJ=\int {\int }_{{[0,1]}^2}\frac{x}{(1+x^2)(1+xy)}dxdy$$$$calculateJbytwomethodsthenfindthevalueofI.$$

Commented by abdo imad last updated on 22/Jan/18

$$wehaveJ={\int }_0^1\left({\int }_0^1\frac{x}{1+xy}dy\right)\frac{dx}{1+x^2}but$$$${\int }_0^1\frac{x}{1+xy}dy={\left[ln(1+xy)\right]}_{y=0}^{y=1}=ln(1+x)so$$$$J={\int }_0^1\frac{ln(1+x)}{1+x^2}dxfromanotherside$$$$J={\int }_0^1\left({\int }_0^1\frac{x}{(1+x^2)(1+xy)}dx\right)dyletdecompose$$$$F\left(x\right)=\frac{x}{(1+x^2)(1+xy)}=\frac{\alpha }{1+xy}+\frac{ax+b}{1+x^2}$$$$\alpha ={lim}_{x\to \frac{-1}{y}}(1+xy)F\left(x\right)=\frac{-1}{y(1+\frac{1}{y^2})}=\frac{-1}{y+\frac{1}{y}}=\frac{-y}{1+y^2}$$$${lim}_{x\to +\propto }xF\left(x\right)=0=\frac{\alpha }{y}+a\Rightarrow a=-\frac{\alpha }{y}=\frac{1}{1+y^2}$$$$F\left(x\right)=\frac{-y}{(1+y^2)(1+xy)}+\frac{\frac{1}{1+y^2}x+b}{1+x^2}$$$$F\left(0\right)=0=\frac{-y}{1+y^2}+b\Rightarrow b=\frac{y}{1+y^2}and$$$$F\left(x\right)=\frac{-y}{(1+y^2)(1+xy)}+\frac{1}{1+y^2(}\frac{x+y}{1+x^2)}$$$${\int }_0^{1}F\left(x\right)dx=\frac{-1}{1+y^2}{\int }_0^1\frac{ydx}{1+xy}+\frac{1}{2(1+y^2)}{\int }_0^1\frac{2x}{1+x^2}dx+\frac{y}{1+y^2}{\int }_0^1\frac{dx}{1+x^2}$$$$=\frac{-1}{1+y^2}{\left[ln(1+xy)\right]}_{x=0}^{x=1}+\frac{1}{2(1+y^2)}{\left[ln(1+x^2)\right]}_0^1$$$$+\frac{\pi }{4}\frac{y}{1+y^2}$$$$-\frac{ln(1+y)}{1+y^2}+\frac{ln2}{2(1+y^2)}+\frac{\pi y}{4(1+y^2)}$$$$J=-{\int }_0^1\frac{ln(1+y)}{1+y^2}dy+\frac{ln2}{2}{\int }_0^1\frac{dy}{1+y^2}+\frac{\pi }{8}{\int }_0^1\frac{2ydy}{1+y^2}$$$$2J=\frac{\pi }{8}ln2+\frac{\pi }{8}ln2=\frac{\pi }{4}ln2\Rightarrow J=\frac{\pi }{8}ln2.$$$$$$$$$$$$$$$$$$

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