I don't have an issue with any of it, but this one question is totally messing with my head, i get different answers by equating coefficients and by letting x=-1. Just totally messing with me
I don't have an issue with any of it, but this one question is totally messing with my head, i get different answers by equating coefficients and by letting x=-1. Just totally messing with me
Partial fractions should be expressed where the degree of the numerator is always less than the degree of the denominator. It is effectively a polynomial division. If the degree of the numerator is greater than or equal to the degree of the denominator then we can reduce it further by polynomial division until the degree of the remainder term falls below the degree of the divisor.
In the case of a quadratic factor (degree 2) which has no real roots (i.e. cannot be factorised further under real numbers), the numerator must be either degree 0 (i.e. a constant) or degree 1. Since we can't be sure exactly which degree the numerator is at first, we always set it to the highest degree (i.e. bx + c which is degree 1) for generalisation and should it actually be a degree zero then the working gives b = 0 for us.
Trebla are you sure? i Got a= 3 b=1 and c=-3 which works out completely. WHat you do is use the substitution method to work out A and then use the Equation coefficients method to work out b and c by inspection?
Answer comes out to be 3ln(x+1) +(1/2)(ln(x^2+1)) -3 tan^-1 (x) + C
Trebla are you sure? i Got a= 3 b=1 and c=-3 which works out completely. WHat you do is use the substitution method to work out A and then use the Equation coefficients method to work out b and c by inspection?
Answer comes out to be 3ln(x+1) +(1/2)(ln(x^2+1)) -3 tan^-1 (x) + C