Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 17

int_1^2 (4y^2-7y-12)/(y(y+2)(y-3))dy
To solve this, apply partial fraction decomposition.
When converting the integrand to sum of proper rational expressions, set the equation as follows:
(4y^2-7y-12)/(y(y+2)(y-3))= A/y + B/(y+2)+C/(y-3)
Multiply both sides by the LCD.
4y^2-7y-12=A(y+2)(y-3)+By(y-3)+Cy(y+2)
4y^2-7y-12=Ay^2-Ay-6A+By^2-3By+Cy^2+2Cy
4y^2-7y-12=(A+B+C)y^2+(-A-3B+2C)Y-6A
For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the polynomials equal to each other.
y^2:
4=A+B+C (Let this be EQ1.)
y:
-7=-A-3B+2C (Let this be EQ2.)
Constant:
-12=-6A (Let this be EQ3.)
To solve for the value of A, consider EQ3.
-12=-6A
2=A
Plug-in this value of A to EQ1.
4=A+B+C
4=2+B+C
2=B+C
Then, isolate the C.
2-B=C
Plug-in this expression and the value of A to EQ2.
-7=-A-3B+2C
-7=-2-3B+2(2-B)
-7=-2-3B+4-2B
-7=2-5B
-9=-5B
9/5=B
And plug-in the value of A and B to EQ1.
4=A+B+C
4=2+9/5+C
4=19/5+C
1/5=C
So the partial fraction decomposition of the integrand is:
(4y^2-7y-12)/(y(y+2)(y-3))= 2/y + (9/5)/(y+2)+(1/5)/(y-3)=2/y+9/(5(y+2))+1/(5(y-3))
Taking the integral of this result to:
int_1^2 (4y^2-7y-12)/(y(y+2)(y-3))dy
=int_1^2 (2/y+9/(5(y+2))+1/(5(y-3)))dy
= 2int_1^2 1/ydy + 9/5int_1^2 1/(y+2)dy+1/5int_1^2 1/(y-3)dy
=(2ln|y| + 9/5ln|y+2| + 1/5ln|y-3|)|_1^2
= (2ln|2| +9/5ln|2+2| +1/5ln|2-3|)-(2ln|1| + 9/5ln|1+2|+1/5ln|1-3|)
=(2ln|2|+9/5ln|4|+1/5ln|-1|) - (2ln|1|+9/5ln|3|+1/5ln|-2|)
=(2ln2+9/5ln2^2+1/5ln1) - (2ln1+9/5ln3+1/5ln2)
=(2ln2+18/5ln2+0)-(0+9/5ln3+1/5ln2)
=28/5ln2-(9/5ln3+1/5ln2)
=28/5ln2-9/5ln3-1/5ln2
=27/5ln2-9/5ln3
=9/5(3ln2-ln3)
=9/5(ln2^3-ln3)
=9/5(ln8-ln3)
=9/5ln(8/3)
Therefore, int_1^2 (4y^2-7y-12)/(y(y+2)(y-3))dy=9/5ln(8/3) .