Help on Partial Fractions

**Paymemoney** · January 5th 2010, 01:22 AM

Hi
The following questions i am having trouble finding the partial fractions to them:

1) $\frac{x^2+x-1}{(x^2+1)^2}$

2)Can someone tell me if i have done this correctly.

$\frac{x^2-1}{x^2-5x+6}$
Do Long Division and i get this: $1+\frac{-5x+7}{(x-3)(x-2)}$
$\frac{A}{(x-3)}+\frac{B}{(x-2)}$

$A(x-2)+B(x-3)=-5x+7$

$A+B=-5$

$A=-5-B$

$-2A-3B=7$

$-2(-5-B)-3B=7$

$10-B=7$

$B=3$

$A=-5-3$

$A=-8$

therefore: $1-\frac{8}{(x-3)}+\frac{3}{(x-2)}$

now answers says its $1-\frac{3}{(x-2)}+\frac{8}{(x-3)}$ .
i have tried doing this equation many times i don't understand how they got the above answer.

P.S

**dedust** · January 5th 2010, 01:33 AM

Originally Posted by Paymemoney

Hi
The following questions i am having trouble finding the partial fractions to them:

1) $\frac{x^2+x-1}{(x^2+1)^2}$

2)Can someone tell me if i have done this correctly.

$\frac{x^2-1}{x^2-5x+6}$
Do Long Division and i get this: $1+\frac{-5x+7}{(x-3)(x-2)}$ $\leftarrow$ should be $\frac{x^2-1}{x^2-5x+6} = \frac{x^2-5x + 6 + 5x -7}{x^2-5x+6} = 1+\frac{5x-7}{(x-3)(x-2)}$

. . .

**Prove It** · January 5th 2010, 02:02 AM

$\int{\frac{x^2 + x - 1}{(x^2 + 1)^2}\,dx}$ .

Let $x = \tan{\theta}$ so that $dx = \sec^2{\theta}\,d\theta$ .

The integral becomes

$\int{\frac{\tan^2{\theta} + \tan{\theta} - 1}{(\tan^2{\theta} + 1)^2}\,\sec^2{\theta}\,d\theta}$

$= \int{\frac{\tan^2{\theta} + \tan{\theta} - 1}{(\sec^2{\theta})^2}\,\sec^2{\theta}\,d\theta}$

$= \int{\cos^2{\theta}(\tan^2{\theta} + \tan{\theta} - 1)\,d\theta}$

$= \int{\sin^2{\theta} + \sin{\theta}\cos{\theta} - 1\,d\theta}$

$= \int{\sin{\theta}\cos{\theta} - \cos^2{\theta}\,d\theta}$

$= \int{\frac{1}{2}\sin{(2\theta)} - \frac{1}{2}\cos{(2\theta)} - \frac{1}{2}\,d\theta}$

$= -\frac{1}{4}\cos{(2\theta)} - \frac{1}{4}\sin{(2\theta)} - \frac{1}{2}\theta + C$

$= -\frac{1}{4}\left(\frac{1 - \tan^2{\theta}}{1 + \tan^2{\theta}}\right) - \frac{1}{4}\left(\frac{2\tan{\theta}}{1 + \tan^2{\theta}}\right) - \frac{1}{2}\theta + C$

$= -\frac{1}{4}\left(\frac{1 - x^2}{1 + x^2}\right) - \frac{1}{4}\left(\frac{2x}{1 + x^2}\right) - \frac{1}{2}\arctan{x} + C$

$= \frac{x^2 - 2x - 1}{4(1 + x^2)} - \frac{1}{2}\arctan{x} + C$ .

**Paymemoney** · January 5th 2010, 02:07 AM

for the first question i want to know how to find the partial fraction not intergrating it.

**Paymemoney** · January 5th 2010, 02:12 AM

Originally Posted by dedust

. . .

ok i understand now, but is there a way of knowing which value you divide by? In this question i divided top into bottom and it was wrong, however when you divide bottom into top you get 5x-7.

**HallsofIvy** · January 5th 2010, 03:00 AM

Originally Posted by Paymemoney

ok i understand now, but is there a way of knowing which value you divide by? In this question i divided top into bottom and it was wrong, however when you divide bottom into top you get 5x-7.

When did you ever learn to divide the denominator of a fraction by the numerator? $\frac{a}{b}$ means a divided by b! You are correct that Paymethemoney's division is wrong. It should be $1+ \frac{5x- 7}{x^2- 5x+ 1}$ . He has the signs in the remainder wrong.

The point in the second problem is that you only use partial fractions on "proper fractions". That's why you need to divide first.

For the first problem, $\frac{x^2+ x- 1}{(x^2+ 1)^2}$ , because the denominator is an irreducible quadratic, squared, we need
$\frac{x^2+ x- 1}{(x^2+ 1)^2}= \frac{Ax+ B}{x^2+ 1}+ \frac{Cx+ D}{(x^2+1)^2}$ .

**dedust** · January 5th 2010, 03:06 AM

Originally Posted by HallsofIvy

It should be $1+ \frac{5x- 7}{x^2- 5x+ 1}$ .

a small correction in the denominator

$1+ \frac{5x- 7}{x^2- 5x+ 6}$

**Paymemoney** · January 5th 2010, 02:56 PM

well this is how i done my long division by divide $x^2-1$ into $x^2-5x+6$

someone tell me where i have gone wrong:

P.S

**ANDS!** · January 5th 2010, 03:24 PM

The only difference is a distributed -1.

**dedust** · January 5th 2010, 03:33 PM

Originally Posted by Paymemoney

well this is how i done my long division by divide $x^2-1$ into $x^2-5x+6$

someone tell me where i have gone wrong:

P.S

nothing wrong,..

but this is $\frac{x^2 - 5x + 6}{x^2 - 1}$

**Paymemoney** · January 6th 2010, 01:40 AM

Originally Posted by HallsofIvy

For the first problem, $\frac{x^2+ x- 1}{(x^2+ 1)^2}$ , because the denominator is an irreducible quadratic, squared, we need
$\frac{x^2+ x- 1}{(x^2+ 1)^2}= \frac{Ax+ B}{x^2+ 1}+ \frac{Cx+ D}{(x^2+1)^2}$ .

i tired doing this method however when i found the values it did not match the book's answers.
Here's what i have done:

$\frac{x^2+x-1}{(x^2+1)^2}$

$\frac{Ax+B}{(x^2+1)}+\frac{Cx+D}{(x^2+1)^2}$

$(Ax+B)(x^2+1)(x^2+1)+(Cx+D)(x^2+1)=x^2+x-1$

The coefficients
$A+B=0$
$2A+C=0$
$2B+D=1$
$A+C=1$
$B+D=-1$

Find B using simultaneous equations:
(1) $2B+D=1$
(2) $B+D=-1$
(1)-(2)=
$B=2$

Find D
$B+D=-1$
$2+D=-1$
$D=-3$

Find A
$A+B=0$
$A+2=0$
$A=-2$

Find C
$A+C=1$
$-2+C=1$
$C=3$
The book's answer is $\frac{1}{x^2+1}+\frac{x-2}{(x^2+1)^2}$

**dedust** · January 6th 2010, 02:51 AM

Originally Posted by Paymemoney

i tired doing this method however when i found the values it did not match the book's answers.
Here's what i have done:

$\frac{x^2+x-1}{(x^2+1)^2}$

$\frac{Ax+B}{(x^2+1)}+\frac{Cx+D}{(x^2+1)^2}$

$(Ax+B)(x^2+1)(x^2+1)+(Cx+D)(x^2+1)=x^2+x-1$ $\rightarrow$ should be $(Ax+B)(x^2+1)+(Cx+D)=x^2+x-1$

. . .

**Prove It** · January 6th 2010, 03:00 AM

Originally Posted by Paymemoney

for the first question i want to know how to find the partial fraction not intergrating it.

Then you should have put this in Algebra or Pre-Calculus.

The only reason I can think of to use Partial Fractions in Calculus would be to find the integral.

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