Mathematics

Resolve into partial fraction $$\displaystyle \frac{7+x}{(1+x)(1+x^2)}$$


ANSWER

$$\displaystyle \frac{3}{1+x}+\frac{4-3x}{1+x^2}$$


SOLUTION
$$\displaystyle \frac { 7+x }{ (1+x)(1+x^{ 2 }) } =\frac { A }{ 1+x } +\frac { Bx+C }{ 1+{ x }^{ 2 } } $$
$$\Rightarrow 7+x=A\left( 1+{ x }^{ 2 } \right) +\left( Bx+C \right) \left( 1+x \right) \\ \Rightarrow 7+x=A\left( 1+{ x }^{ 2 } \right) +B\left( x+{ x }^{ 2 } \right) +C\left( 1+x \right) $$
On comparing coefficients 
$$A+B=0,B+C=1,A+C=7\Rightarrow A=3,B=-3,C=4$$
Hence, option 'C' is correct.
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Single Correct Medium Published on 17th 09, 2020
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