Question

resolve into partial fraction 3x+1/(x-3)(x+1)​

Answer 1

Step-by-step explanation:

$\frac{3x + 1}{(x - 2)(x + 1)} = \frac{a}{(x - 2)} + \frac{b}{(x + 1)}$

$\frac{3x + 1}{(x - 2)(x + 1)} = \frac{a(x + 1) + b(x - 2)}{(x - 2)(x + 1)}$

$(3x + 1) = a(x + 1) + b(x - 2)$

Putting x=2

$(3 \times 2 + 1) = a(2 + 1) + b(2 - 2) \\ = > 7 = 6a \\ = > a = \frac{7}{6}$

Putting x=1

$3 \times (- 1 )+ 1 = a( - 1 + 1) + b( - 1 - 2)$

$- 2 = - 3b$

$b = \frac{2}{3}$

Answer 2

Given,

$\[\frac{{3x + 1}}{{\left( {x - 3} \right)\left( {x + 1} \right)}}\]$

Solution,

Calculate the partial fraction,

$\[\frac{{3x + 1}}{{\left( {x - 3} \right)\left( {x + 1} \right)}} = \frac{A}{{\left( {x - 3} \right)}} + \frac{B}{{\left( {x + 1} \right)}}\]$

$\[ \Rightarrow \frac{{3x + 1}}{{\left( {x - 3} \right)\left( {x + 1} \right)}} = \frac{{A\left( {x + 1} \right) + B\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 1} \right)}}\]$

$\[ \Rightarrow \frac{{3x + 1}}{{\left( {x - 3} \right)\left( {x + 1} \right)}} = \frac{{Ax + A + Bx - 3B}}{{\left( {x - 3} \right)\left( {x + 1} \right)}}\]$

$\[ \Rightarrow 3x + 1 = \left( {A + B} \right)x + \left( {A - 3B} \right)\]$

Compare both equations,

$\[\begin{array}{l}A + B = 3 - - - - \left( 1 \right)\\A - 3B = 1 - - - - \left( 2 \right)\end{array}\]$

Solve both equations,

$\[A = \frac{5}{2},B = \frac{1}{2}\]$

Partial fraction $\[ = \frac{5}{{2\left( {x - 3} \right)}} + \frac{1}{{2\left( {x + 1} \right)}}\]$

Hence the partial fraction is $\[\frac{5}{{2\left( {x - 3} \right)}} + \frac{1}{{2\left( {x + 1} \right)}}\]$.