Question

could someone help me in how to solve partial fractions
Integrate(4w^2+3w^2+9/w^8-9w)dw

Answer 1

First factor the denominator.

Answer 2

ok then i'm sorry could please explain it to with some more detail thanks

Answer 3

i really don't know how to do this for me its like from scratch

Answer 4

\[
\frac{4w^2+3w^2+9}{w^8-9w}=\frac{7w^2+9}{w(w^7-9)}
\]

Answer 5

Each denominator gets its own fraction: \[
\frac{7w^2+9}{w(w^7-9)} = \frac{A}{w}+\frac{B}{w^7-9}
\]

Answer 6

how did you got 7w^2+9 for nominator?

Answer 7

Then you try to add the fractions as are: \[
\frac{7w^2+9}{w(w^7-9)} = \frac{A}{w}+\frac{B}{w^7-9}=\frac{A(w^7-9)+Bw}{w(w^7-9)}
\]

Answer 8

You wrote:\[
4w^2+3w^2+9
\]Add like terms: \[
7w^2+9
\]

Answer 9

i'm sorry i made a mistake but 3w not 3w^2

Answer 10

but i got it the point

Answer 11

In this case fraction decomposition won't work because there is no solution for \(A\) and \(B\).

Answer 12

ok

Answer 13

Unless you wrote it incorrectly...

Answer 14

no i wrote it correctly but can't we solve for A and B?

Answer 15

Doesn't look like it: \[
4w^2+2w+9=Aw^7+Bw-9A
\]

Answer 16

yeah i guess it wouldn't work like this

Answer 17

Shouldn't it be

\(\color{blue}{\text{Originally Posted by}}\) @wio
\[\frac{7w^2+9}{w(w^7-9)} = \frac{A}{w}+\frac{\color{red}{Bw^6+Cw^5+Dw^4+Ew^3+Fw^2+Gw+H}}{w^7-9}~~?
\]\(\color{blue}{\text{End of Quote}}\)

Answer 18

A lot of those constants will likely disappear, I think.

Answer 19

The \(Bx+C\) comes when you have denominators with multiplicity greater than 1.

Answer 20

Right, but if you have a quadratic factor in the denominator, the numerator of the partial fraction with that denominator will be a linear factor.

Answer 21

So in general, if you have an \(n\)-th degree factor in the denominator, you should have an \(n-1\)-th degree factor in the numerator. In the case of greater-than-1 multiplicity, you can use \(Bx+C\) for a factor with mult. 2, but you'll find that \(B=0\).

Answer 22

http://www.wolframalpha.com/input/?i=Integrate%28%284w%5E2%2B3w%5E2%2B9%29%2F%28w%5E8-9w%29dw

no other short way ?

Answer 23

\[\begin{align*}\frac{4w^2+3w+9}{w(w^7-9)} &= \frac{A}{w}+\frac{Bw^6+Cw^5+Dw^4+Ew^3+Fw^2+Gw+H}{w^7-9}\\
4w^2+3w+9&=Aw^7-9A+Bw^7+Cw^6+Dw^5+Ew^4+Fw^3+Gw^2+Hw\\
&=(A+B)w^7+Cw^6+Dw^5+Ew^4+Fw^3+Gw^2+Hw-9A
\end{align*}\]
which gives the system
\[\begin{cases}
A+B=0\\C=0\\D=0\\E=0\\F=0\\G=4\\H=3\\-9A=9
\end{cases}~~\Rightarrow~~A=-1,~B=1\]