Question

I'm having a hard time understanding partial fraction decomposition.

\[\int\limits_{1}^{2}\frac{ x^3 + 4x^2 +x -1 }{ x^3 + x^2 }dx\]

Answer 1

if degs match or if degree is greater on top you must do division first

Answer 2

Ok so after I get the division I got:
\[1+\frac{ 3x^2+x-1 }{ x^3+x^2 }\]

Answer 3

Did*

Answer 4

ok partial fraction time
first step factor denominator

Answer 5

Ok, I know it can be factored into: (x^2)(x + 1)
Not sure if I can factor it any more than that

Answer 6

ok cool and no

Answer 7

so x^2 is a linear factor^2 and x+1 is just a linear factor non-repeated

Answer 8

\[\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}=\frac{3x^2+x-1}{x^3+x^2}\]

Answer 9

Oh ok. I wasn't sure we were supposed to have A/x and then B/x^2. I thought there wouldn't be a coefficient over x.

Answer 10

I always do constant/linear^n
1 degree less than the function inside the power on the bottom

Answer 11

Ok that makes sense!
So afterwards we would multiply everything by the denominator?

Answer 12

well we find the lcm which is x^2(x+1)

use this to combine the fractions

Answer 13

example that first fraction I wrote we multiply top and bottom by x(x+1)
second fraction we do (x+1)
last fraction x^2

Answer 14

\[\frac{Ax(x+1)+B(x+1)+Cx^2}{x^2(x+1)}=\frac{3x^2+x-1}{x^3+x^2} \\ \implies Ax(x+1)+B(x+1)+Cx^2=3x^2+x-1 \\ \text{ now I have started \to really like heaviside method for this }\]

Answer 15

Ok I understand that part. Do we distribute next?

Answer 16

you can ... or use heaviside method...( I think you will like this more)

Answer 17

Since this is an equation that means we want both sides to be the same for any x

Answer 18

this means this equation should hold
that is left side should equal right side when x=0 or when x=-1 ....and so on...

Answer 19

So enter in 0 for x on both sides you should get an easy equation to obtain B

Answer 20

Oh so B would end up being equal to -1?

Answer 21

yep now select x=-1
I'm choosing x=-1 because a lot of that stuff on the left hand side will disappear

Answer 22

Ok so c=1?

Answer 23

yep ok... now choose any value for x (an easy one preferably) and input your values for B and C into that equation and solve for A

Answer 24

any x you haven't used already*

Answer 25

like x=1

Answer 26

it doesn't matter which value for x you choose those you should wind up with A=2 (if I didn't make a mistake lol)

Answer 27

Yeah I got A=2 too when I chose x=2

Answer 28

I have to get ready to go out
but I hope that helps
@IrishBoy123 mentioning you because I have to leave

Answer 29

Ok thank you for the help! This way is much better!

Answer 30

good night @freckles !