# Math Help - Help on Partial Fractions

1. ## Help on Partial Fractions

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

2. 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 $\frac{x^2-1}{x^2-5x+6} = \frac{x^2-5x + 6 + 5x -7}{x^2-5x+6} = 1+\frac{5x-7}{(x-3)(x-2)}$
. . .

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

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

The integral becomes

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

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

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

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

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

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

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

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

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

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

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

5. 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.

6. 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
$\frac{x^2+ x- 1}{(x^2+ 1)^2}= \frac{Ax+ B}{x^2+ 1}+ \frac{Cx+ D}{(x^2+1)^2}$.

7. Originally Posted by HallsofIvy
It should be $1+ \frac{5x- 7}{x^2- 5x+ 1}$.
a small correction in the denominator

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

8. 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

9. The only difference is a distributed -1.

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

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

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

13. 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.