How to Solve Partial Fraction Decomposition for (2x+3)/(x+1)^2

[Rasine]

(2x+3)/(x+1)^2


so this is what i am thinking...but it does not make sense


=(A/x+1)+(B/(x+1)^2)

so then 2x+3=A(x+1)^2+B(x+1)

2x+3=Ax^2+A2x+A+Bx+1

so that would make...
0=A
2=2A+B
3=A+B

this solution does not make any sense becuase if A=0 then according to the second equation B=2 which is not what B would equal in the third equation.

what am i doing wrong?

 

[Dick]

Your algebra is wrong

2x+3=A*(x+1)+B is correct.

I'm not sure how you are getting the equation you are working with.

 

[Rasine]

what happened to to (x+1)^2 term?

 

[cristo]

so this is what i am thinking...but it does not make sense


(2x+3)/(x+1)^2=(A/x+1)+(B/(x+1)^2) (*)

so then 2x+3=A(x+1)^2+B(x+1)

This is wrong. Multipying both sides of (*) by (x+1)^2 gives 2x+3=A(x+1)+B

 

[Dick]

Multiplying both sides by (x+1)^2 cancels one (x+1) from the A term and both from the B term.

 

[Rasine]

so should it be 2x+3/(x+1)(x+1)=(A/x+1)+(B/x+1)

here..i just split up the (x+1)^2


this is right?

 

[Dick]

(2x+3)/(x+1)^2=A/(x+1)+B/(x+1)^2. CAREFULLY multiply each of those three terms by (x+1)^2 and report back the results.

 

[Rasine]

ohhh ok...i think i understand..

so if i have (2x+3)/(x+1)^2=(A/x+1)+(B/(x+1)^2) (*)

then 2x+3=A(x+1)+B because i multiply both sides by (x+1)^2

 

[Rasine]

thank you very much