We have to find the partial fractions of (5x+1)/(x^2+x-2)
First let's factorize the denominator
x^2 + x - 2
=> x^2 + 2x - x - 2
=> x(x + 2) - 1(x + 2)
=> (x - 1)(x + 2)
(5x+1)/(x^2+x-2) = (5x+1)/(x - 1)(x + 2)
The partial fractions are A /(x - 1) + B/(x + 2)
A /(x - 1) + B/(x + 2) = (5x+1)/(x - 1)(x + 2)
=> A(x + 2) + B(x - 1) = 5x + 1
=> Ax + 2A + Bx - B = 5x + 1
=> 2A - B = 1 and A + B = 5
Add the two equations
=> 3A = 6
=> A = 2
B = 5 - A = 3
The partial fractions of (5x+1)/(x^2+x-2) are 2/(x - 1) + 3/(x + 2)