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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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