Mathematics

Resolve into partial fractions $$\displaystyle \frac{4x^3-7x^2+5x-4}{(x^2+1)(x^2-3x+2)}$$


ANSWER

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


SOLUTION
Let $$\displaystyle \frac { 4x^{ 3 }-7x^{ 2 }+5x-4 }{ \left( x^{ 2 }+1 \right) \left( x^{ 2 }-3x+2 \right)  } =\frac { Ax+B }{ { x }^{ 2 }+1 } +\frac { C }{ x-2 } +\frac { D }{ x-1 } $$
$$\Rightarrow 4x^{ 3 }-7x^{ 2 }+5x-4=\left( Ax+B \right) \left( x-2 \right) \left( x-1 \right) +C\left( { x }^{ 2 }+1 \right) \left( x-1 \right) +D\left( { x }^{ 2 }+1 \right) \left( x-2 \right) \\ \Rightarrow 4x^{ 3 }-7x^{ 2 }+5x-4=A\left( { x }^{ 3 }-3{ x }^{ 2 }+2x \right) +B\left( { x }^{ 2 }-3x+2 \right) +C\left( { x }^{ 3 }-{ x }^{ 2 }-x-1 \right) +D\left( { x }^{ 3 }-2{ x }^{ 2 }+x-2 \right) $$
On comparing coefficients we get
$$A+C+D=4,-3A+B-C-2D=-7,2A-3B-C+D=5,2B-C-2D=-4\\ \Rightarrow A=1,B=0,C=2,D=1$$
Hence
$$\displaystyle \frac { 4x^{ 3 }-7x^{ 2 }+5x-4 }{ \left( x^{ 2 }+1 \right) \left( x^{ 2 }-3x+2 \right)  } =\frac { x }{ { x }^{ 2 }+1 } +\frac { 2 }{ x-2 } +\frac { 1 }{ x-1 } $$
Hence, option 'C' is correct.
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