SOLUTION: how do you do partial fraction decomposition?? Like on a problem such as: 1/(x^2+x)???

Question 65100: how do you do partial fraction decomposition?? Like on a problem such as: 1/(x^2+x)???
Found 2 solutions by venugopalramana, FadelSalem:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
1/(X^2+X)
1.FACTORISE D.R
X^2+X=X(X+1)
2.WRITE THE GIVEN EXPRESSION AS A SUM OF TERMS WITH EACH OF THE FACTORS IN D.R.
IF D.R IS DEGREE 1 PUT CONSTANTS LIKE A OR B OR C ETC IN THE N.R
IF D.R IS SECOND DEGREE PUT FIRST DEGREE BINOMIAL IN NR LIKE AX+B ETC
IF DR IS THIRD DEGREE PUT SECOND DEGREE TRINOMIAL IN NR..LIKE AX^2+BX+C ETC..
3. HERE D.R IS I DEGREE HENCE
1/(X^2+X) = [A/X] + [B/(X+1)]
SIMPLIFY RHS ...WE GET
1/X(X+1) = [A(X+1)+BX]/X(X+1)
EQUATE NRS AS DRS ARE SAME
1= A(X+1)+BX
GIVE DIFFERENT VALUES TO X ON LHS AND RHS TO GET EQNS TO SOLVE FOR A & B
GIVE X=0
1=A(0+1)=A...SO.A=1
GIVE X=-1
1=-B
B=-1
SO
[1/X(X+1)] = [1/X]- [1/(X+1)]
Answer by FadelSalem(12) (Show Source): You can put this solution on YOUR website!
Hey, the answer is very easy as follwos:
1/(x^2+x) = 1/ x(x+1) = A/x + B/(x+1)
Multiply each side by x(x+1), we'll get:
1 = A(x+1) + B x
1 = Ax + A + Bx
Compare absolute terms we get: 1 = A
compare x coefficients we get: 0 = A + B [and that means B = -1]
So, 1/(x^2+x) = 1/x - 1/(x+1)
you're welcome any time and wish you Happy new year
Fadel Salem - Egypt
FadelSalem@hotmail.com
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