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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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