Mathematics

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


SOLUTION
$$\displaystyle\int { \dfrac { 2 }{ (1-x)(1+{ x }^{ 2 }) }  } =$$ $$\displaystyle\int { \dfrac { A }{ (1-x) }  } $$$$+\displaystyle\int { \dfrac { Bx+C }{(1+{ x }^{ 2 }) }  } $$

$$2=A(1+x^2)+(Bx+C)(1-x)$$      .........(1)

put $$x=1$$ in eq.1

$$2=A(1+1^2)+(B(1)+C)(1-1)$$

$$2A=2$$

$$A=1$$

and now put $$x=0 and A=1$$ in eq. (1)

$$2=1(1+0)+(B(0)+C)(1-0)$$

$$C=1$$

put value of $$C$$ and $$A$$ and $$x=-1$$  in eq. (1)

$$2=(1)(1+(-1)^2)+(B(-1)+1)(1-(-1))$$

$$2=2+2(1-B)$$

$$B=1$$

now,
$$\displaystyle\int { \dfrac { 2 }{ (1-x)(1+{ x }^{ 2 }) }  } =$$ $$\displaystyle\int { \dfrac { 1 }{ (1-x) }  } $$$$+\displaystyle\int { \dfrac { x+1 }{(1+{ x }^{ 2 }) }  } $$

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

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

$$=$$ $$-log(1-x)+\dfrac {1}{2} log(1+x^2)+$$ $$\tan ^{ -1 }{ x } +C$$
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Subjective Medium Published on 17th 09, 2020
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