Decomposing Partial Fraction

[r2k3982]

1+lnx[sup:23y2a125]2[/sup:23y2a125]/(lnx+2)(lnx-3)[sup:23y2a125]2[/sup:23y2a125]

Is it right to substitute a variable (i picked y) for lnx?

I did that and then got:

A/(y+2) + B/(y-3) + C/(y-3)[sup:23y2a125]2[/sup:23y2a125]

Then I multiplied by the common denominator (y+2)(y-3)[sup:23y2a125]2[/sup:23y2a125]

and I get:
A(y-3)[sup:23y2a125]2[/sup:23y2a125] + B(y+2)(y-3) + C(y+2)

I FOILed
A(y[sup:23y2a125]2[/sup:23y2a125]-6y+9) + B(y[sup:23y2a125]2[/sup:23y2a125]-y-6) + C(y+2)

Distribute A,B,C

Ay[sup:23y2a125]2[/sup:23y2a125] -6Ay+9A + By[sup:23y2a125]2[/sup:23y2a125]-By-6B + Cy+2C

Then I grouped all the like terms.

(A+B)y[sup:23y2a125]2[/sup:23y2a125] + (-6A-B+C)y + (9A-6B+2C)


okay here is where I am confused. In the numerator, it has lnx[sup:23y2a125]2[/sup:23y2a125], but my teacher said to write it as 2lnx. But when you are going to figure out what A,B,and C do you use

1=(A+B) since it would be 1y[sup:23y2a125]2[/sup:23y2a125] OR
2=(-6A-B+C) because I moved the 2 in front of the lnx making it 2y?

I did the latter

and came up with values of A=-3/25, B=3/25, C=7/5

 

[mmm4444bot]

1=(A+B) since it would be 1y[sup:glltoznq]2[/sup:glltoznq] No.

ln(x^2) is not the same as ln(x)^2

0 = A + B

2=(-6A-B+C) because I moved the 2 in front of the lnx making it 2y? Yes, the coefficient on y^1 is 2.

0y^2 + 2y + 1 = (A + B)y2 + (-6A - B + C)y + (9A - 6B + 2C)

I did the latter

and came up with values of A=-3/25, B=3/25, C=7/5 This looks correct, to me.