Mathematics

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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 126

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