Question

Ressolve on partial fraction


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Answer 1

Answer:

Step-by-step explanation:

let

   $\frac{3x+1}{(x-2)(x+1)} = \frac{A}{x-2} +\frac{B}{x+1}$

i.e. $\frac{3x+1}{(x-2)(x-1)} = \frac{A(x+1)+B(x-2)}{(x-2)(x+1)}$

equating numerator parts

$3x+1=A(x+1)+B(x-2)$

this equation is true for any value of x

to find A and B

put x = -1

$-3+1=A(0)+B(-1-2)\\-3B=-2\\B=\frac{2}{3}$

put x = 2

$3(2)+1=A(2+1)+B(0)\\3A=7\\A=\frac{7}{3}$

$hence, \frac{(3x+1)}{(x-2)(x+1)}$

$=\frac{7}{3(x-2)} +\frac{2}{3(x+1)}$