Partial Fraction Decomp. Integral

[3.141592654]

Homework Statement

\int\frac{4y^2-7y-12}{y(y+2)(y-3)} with limits of integration [1, 2]

Homework Equations

The answer given in the book is \frac{27}{5}ln(2)-\frac{9}{5}ln(3)

or \frac{9}{5}ln\frac{8}{3}

The Attempt at a Solution

\int\frac{4y^2-7y-12}{y(y+2)(y-3)}

=\int\frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-3}

I found:

4y^2-7y-12=A(y+2)(y-3)+B(y)(y-3)+C(y)(y+2)

If y=-2, then B=1

I also found:

4y^2-7y-12=(A+B+C)y^2+(2C-3B-A)y-6A

so 6A=12, or A=2.

Also, (A+B+C)=4 so (2+1+C)=4 and C=1

Therefore: =\int\frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-3}

=\int\frac{2}{y}+\frac{1}{y+2}+\frac{1}{y-3}

=2ln|y|+ln|y+2|+ln|y-3| with limits [1, 2]

=2(ln(2)-ln(1))+(ln(2+2)-ln(1+2)+(ln|2-3|-ln|1-3|)

=2ln(2)+(ln(4)-ln(3)+(ln(1)-ln(2))

=2ln(2)+ln(\frac{4}{3})-ln(2)

This is not equal to either answer provided in the book but I can't find an error. Also, how do you put the limits of integration into the function using Latex?

 

[Mark44]

Check your work. I get A = 2, B = 9/5, and C = 1/5.

 

[Dick]

How do you get B=1? I get B=9/5. Also note you can get A (by putting y=0) and C (by putting y=3) the same way you got B.