0

How to solve this using partial fraction $$\frac{x^2+7}{(2x-1)(x-1)}$$

I am using $\frac{A}{2x-1}+\frac{B}{x-1}$ but not getting it

Ángel Mario Gallegos
  • 19,882
Milan Amrut Joshi
  • 1,109
  • 2
    You have to first perform division. You can only split into partial fractions if the enumerator is of lower degree than denominator. – 5xum Mar 03 '16 at 07:50

2 Answers2

1

First perform long division: \begin{align} \frac{x^2+7}{(2x-1)(x-1)}&=\frac{1}{2}+\frac{\frac{3}{2}x+\frac{13}{2}}{(2x-1)(x-1)} \end{align} Then, put $$\frac{\frac{3}{2}x+\frac{13}{2}}{(2x-1)(x-1)}=\frac{A}{2x-1}+\frac{B}{x-1}$$

Ángel Mario Gallegos
  • 19,882
0

Hint: Subtract $-8$ and add back $8$ to $x^2+7$.

Salech Alhasov
  • 6,780
  • 2
  • 29
  • 47