Solving Partial Fractions for Integrals

[TsAmE]

Homework Statement

Evaluate the integral:

\int_{1}^2 \frac{4y^2 - 7y -12}{y(y+2)(y-3)}dy<br />

Homework Equations

None.

The Attempt at a Solution

\frac{4y^2 -7y -12}{y(y+2)(y-3)} = \frac{A}{y} + \frac{B}{y+2} + \frac{D}{y-3}

y = 0: -12 = -6A \rightarrow A=2

y = -2: 16 + 14 - 12 = 10B

18 = 10B \rightarrow B = \frac{18}{10} = \frac{9}{5}

y = 3: 36 - 21 -12 = 15D<br />

3 = 15D \rightarrow D = \frac{3}{15} = \frac{1}{5}

=\int_{1}^2 \frac{2}{y} + \frac {\frac{9}{5}}{y + 2} + \frac{\frac{1}{5}}{y - 3} dy

=2ln|y| + 9ln|5y+2| + ln|y - 3| ]_{1}^{2}

2ln2 + 9ln12 - (9ln7 + ln2)

ln2 + 9ln12 - 9ln7<br /> <br />

The correct answer is \frac {27}{5}ln2 - \frac{9}{5}ln3 but I can't see what I did wrong.

 

[Mark44]

Homework Statement

Evaluate the integral:

\int_{1}^2 \frac{4y^2 - 7y -12}{y(y+2)(y-3)}dy<br />

Homework Equations

None.

The Attempt at a Solution

\frac{4y^2 -7y -12}{y(y+2)(y-3)} = \frac{A}{y} + \frac{B}{y+2} + \frac{D}{y-3}

y = 0: -12 = -6A \rightarrow A=2

y = -2: 16 + 14 - 12 = 10B

18 = 10B \rightarrow B = \frac{18}{10} = \frac{9}{5}

y = 3: 36 - 21 -12 = 15D<br />

3 = 15D \rightarrow D = \frac{3}{15} = \frac{1}{5}

=\int_{1}^2 \frac{2}{y} + \frac {\frac{9}{5}}{y + 2} + \frac{\frac{1}{5}}{y - 3} dy

=2ln|y| + 9ln|5y+2| + ln|y - 3| ]_{1}^{2}

The mistakes are above. The 2nd and 3rd terms in your antiderivative are incorrect. I was able to get the same answer as in the book.

2ln2 + 9ln12 - (9ln7 + ln2)

ln2 + 9ln12 - 9ln7<br /> <br />

The correct answer is \frac {27}{5}ln2 - \frac{9}{5}ln3 but I can't see what I did wrong.