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)**