how do you go from dv/(v+2)(v-1) to -(1/3)/(v+2) + (1/3)/(v-1) dv?

Answer 1

You use partial fractions.

1/(v+2)(v-1) = 1/3 (1/(v-1) - 1/(v+2))

Answer 2

for partial fractions don't you do A/(v+2) + B/ (v-1) = (v+2)(v-1)?

how do you go from there?
can you show what you did step by step?

Answer 3

Surely this treated in your text. You want

A/(v-1) + B/(v+2) = 1/(v-1)(v+2)
A(v+2) + B(v-1) = 1
Av + 2A + Bv - B = 1
(A+B)v + (2A-B) = 0v + 1
For those polynomials to be identical, all of the coefficients must match. So, we have

A+B = 0
2A-B = 1
solve those and you get
A = 1/3
B = -1/3

Thus, the solution to the partial fraction breakdown.