Mathematics

Find :
$$\displaystyle\int {\dfrac{2}{(1-x)(1+x^2)}dx}$$


SOLUTION

We have,


$$I=\displaystyle\int{\dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}dx}$$


$$ I=\displaystyle\int{\left( \dfrac{1}{1-x}+\dfrac{x+1}{1+{{x}^{2}}} \right)dx} $$


$$ I=-\displaystyle\int{\left( \dfrac{1}{x-1} \right)dx+\dfrac{1}{2}}\displaystyle\int{\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx+\displaystyle\int{\left( \dfrac{1}{1+{{x}^{2}}} \right)dx}} $$


$$ I=-\ln \left( x-1 \right)+\dfrac{1}{2}\ln \left( {{x}^{2}}+1 \right)+{{\tan }^{-1}}x+C $$


 


Hence, this is the answer.

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