In the given fraction, the denominator has two linear,non-repeated factors.
`:.` The given fraction can be written as the sum of two partial fractions.
Let `(3x+5)/((x+2)(3x-1))=(A)/(x+2)+(B)/(3x-1)`
`implies3x+5=A(3x-1)+B(x+2)` .......`(1)`
Put `x=-2` in Eq. `(1)`,
`implies3(-2)+5=A[3(-2)-1]+B(-2+2)-1=-7AimpliesA=(1)/(7)`
Comparing the costant the constant terms on both sides of Eq. `(1)`, we have
`5=-A+2B`
`5=-(1)/(7)+2B` or `B=(18)/(7)`
`:. (3x+5)/((x+2)(3x-1))=((1)/(7))/(x+2)+((18)/(7))/(3x-1)=(1)/(7)[(1)/(x+2)+(18)/(3x-1)]`
