Asked by sally
                how do you go from dv/(v+2)(v-1) to -(1/3)/(v+2) + (1/3)/(v-1) dv?
            
            
        Answers
                    Answered by
            STEVE
            
    You use partial fractions.
1/(v+2)(v-1) = 1/3 (1/(v-1) - 1/(v+2))
    
1/(v+2)(v-1) = 1/3 (1/(v-1) - 1/(v+2))
                    Answered by
            sally
            
    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?
    
how do you go from there?
can you show what you did step by step?
                    Answered by
            Steve
            
    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.
    
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.
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