Mathematics

Find the partial fraction
$$\dfrac { 2x+1 }{ \left( 3x+2 \right) \left( 4{ x }^{ 2 }+5x+6 \right)  }$$.


SOLUTION
Let $$\dfrac { 2x+1 }{ \left( 3x+2 \right) \left( 4{ x }^{ 2 }+5x+6 \right)  } =\dfrac { A }{ \left( 3x+2 \right)  } \dfrac { Bx+C }{ \left( 4{ x }^{ 2 }+5x+6 \right)  }$$
then $$2x+1=A\left( 4{ x }^{ 2 }+5x+6 \right) +\left( Bx+C \right) \left( 3x+2 \right)$$
where $$A,B,C$$ are constants.
For $$A$$, let   $$3x+2=0$$,
i.e.,       $$x=-2/3$$
$$2\left( -\dfrac { 2 }{ 3 }  \right) +1=A\left\{ 4.\dfrac { 4 }{ 9 } -\dfrac { 10 }{ 3 } +6 \right\} +\left\{ B\left( -\dfrac { 2 }{ 3 }  \right) +C \right\} \left( 0 \right)$$
$$-\dfrac { 1 }{ 3 } =A\left( \dfrac { 40 }{ 9 }  \right) \Rightarrow A=-\dfrac { 3 }{ 40 }$$
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Subjective Medium Published on 17th 09, 2020
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