Need help finishing Impartial Fraction

**mmattson07** · October 6th 2009, 08:24 PM

Hello I'm working on an impartial fraction problem

$\int_0^1 \! \frac {x^3-x-3}{x^2-x-6} \, dx.$

Long dividing we get

$\int_0^1 \! x- \frac {x^2-7x-3}{x^2-x-6} \, dx.$

I don't know if i have to long divide again or do partial fraction exp...

Thanks

**tonio** · October 6th 2009, 08:35 PM

Originally Posted by mmattson07

Hello I'm working on an impartial fraction problem

$\int_0^1 \! \frac {x^3-x-3}{x^2-x-6} \, dx.$

Long dividing we get

$\int_0^1 \! x- \frac {x^2-7x-3}{x^2-x-6} \, dx.$

I don't know if i have to long divide again or do partial fraction exp...

Thanks

The division is incorrect:

[x^3 - x - 3]/[x^2 - x - 6] = x + 1+ [6x + 3] /[x^2 - x - 6] and now do an easy exercise of partial fractions on [6x + 3] /[x^2 - x - 6] = 3(2x + 1)/(x-3)(x+2)

Tonio

**mmattson07** · October 6th 2009, 10:58 PM

Ok so I am stuck again...

Here is my PFE

$\frac{6x+3}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}$

$6x+3=A(x+2)+B(x-3)$

So A= 21/5 and B= -9/5

$\int_0^1 \! \frac{21}{5}(\frac{1}{x-3})+(\frac{9}{5})\frac{1}{x+2} \, dx.$

$\frac{21}{5}ln|x-3|+\frac{9}{5}ln|x+2|\bigg|_0^1$

Im not getting the correct answer...did i make a mistake?

**mr fantastic** · October 7th 2009, 12:26 AM

Originally Posted by mmattson07

Ok so I am stuck again...

Here is my PFE

$\frac{6x+3}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}$

$6x+3=A(x+2)+B(x-3)$

So A= 21/5 and B= -9/5 Mr F says: From what I see below I suppose you meant B = 9/5 rather than -9/5.

$\int_0^1 \! \frac{21}{5}(\frac{1}{x-3})+(\frac{9}{5})\frac{1}{x+2} \, dx.$

$\frac{21}{5}ln|x-3|+\frac{9}{5}ln|x+2|\bigg|_0^1$

Im not getting the correct answer...did i make a mistake?

You have forgotten to include $\int_0^1 x + 1 \, dx$ ....

**mmattson07** · October 7th 2009, 12:40 AM

So it should be

$\frac{3}{2}+\frac{21}{5}ln(-2)+\frac{9}{5}ln(3)-\frac{21}{5}ln(-3)-\frac{9}{5}ln(2)$

Doesnt look like it will simplify to any of these answers

$\frac{1}{2}-\frac{12}{5}ln(\frac{3}{2})$

$\frac{3}{2}-\frac{12}{5}ln(\frac{3}{2})$

$\frac{1}{2}-\frac{12}{5}ln(\frac{2}{3})$

**mr fantastic** · October 7th 2009, 12:54 AM

Originally Posted by mmattson07

So it should be

$\frac{3}{2}+\frac{21}{5}ln(-2)+\frac{9}{5}ln(3)-\frac{21}{5}ln(-3)-\frac{9}{5}ln(2)$ Mr F says: Things like ln(-2) and ln(-3) are horrible and just plain sloppy. Surely you mean ln|-2| and ln|-3|.

Doesnt look like it will simplify to any of these answers

$\frac{1}{2}-\frac{12}{5}ln(\frac{3}{2})$

$\frac{3}{2}-\frac{12}{5}ln(\frac{3}{2})$

$\frac{1}{2}-\frac{12}{5}ln(\frac{2}{3})$

The answer is $\frac{3}{2} + \frac{21}{5} \ln \frac{2}{3} + \frac{9}{5} \ln \frac{3}{2}$ .

Using a simple log rule this can be re-written as $\frac{3}{2} - \frac{21}{5} \ln \frac{3}{2} + \frac{9}{5} \ln \frac{3}{2}$ ....

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