`int_1^2 (4y^2 - 7y - 12)/(y(y + 2)(y - 3)) dy` Evaluate the integral

`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...

`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)` .